Bài 48 (trang 29 SGK Toán 9 Tập 1)
Khử mẫu của biểu thức lấy căn
$\sqrt{\dfrac{1}{600}}; \sqrt{\dfrac{11}{540}}$ ; $\sqrt{\dfrac{3}{50}} ; \sqrt{\dfrac{5}{98}}$ ; $\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}$
Bài 49 (trang 29 SGK Toán 9 Tập 1)
Khử mẫu của biểu thức lấy căn
$ab\sqrt{\dfrac{a}{b}}$ ; $\dfrac{a}{b} \sqrt{\dfrac{b}{a}}$ ; $\sqrt{\dfrac{1}{b}+\dfrac{1}{b^{2}}}$ ; $\sqrt{\dfrac{9 a^{3}}{36 b}}$ ; $3 xy \sqrt{\dfrac{2}{x y}}$.
(Giả thiết các biểu thức có nghĩa).
(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)
#Học tốt!!!
\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
\(a\cdot b\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{a\cdot b}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
Bài 52 (trang 30 SGK Toán 9 Tập 1)
Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa
$\dfrac{2}{\sqrt{6}-\sqrt{5}}$ ; $\dfrac{3}{\sqrt{10}+\sqrt{7}}$ ; $\dfrac{1}{\sqrt{x}-\sqrt{y}}$ ; $\dfrac{2 a b}{\sqrt{a}-\sqrt{b}}$.
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
Bài 51 (trang 30 SGK Toán 9 Tập 1)
Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa
$\dfrac{3}{\sqrt{3}+1}$ ; $\dfrac{2}{\sqrt{3}-1}$; $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}$ ; $\dfrac{b}{3+\sqrt{b}}$ ; $\dfrac{p}{2 .\sqrt{p}-1}$.
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33−33−1=33−32" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.23−1=2(3+1)(3−1)(3+1)=2(3+1)(3)2−12" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
2(3+1)3−1=2(3+1)2=3+1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.2+32−3=(2+3).(2+3)(2−3)(2+3)=(2+3)222−(3)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
22+2.2.3+(3)24−3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
7+431=7+43" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.b3+b=b(3−b)(3+b)(3−b)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
b(3−b)32−(b)2=b(3−b)9−b;(b≠9)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.p2p−1=p(2p+1)(2p−1)(2p+1)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
p(2p+1)(2p)2−12=p(2p+1)4p−1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
#Ye Chi-Lien
\(\frac{3}{\sqrt{3}+1}=\frac{3\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}\)
\(\frac{2}{\sqrt{3}-1}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\sqrt{3}-1\)
\(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3}=\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)
\(\frac{b}{3+\sqrt{b}}=\frac{b\left(3-\sqrt{b}\right)}{\left(3+\sqrt{b}\right)\left(3-\sqrt{b}\right)}=\frac{b\left(3-\sqrt{b}\right)}{9-b}\)
\(\frac{p}{2\sqrt{p}-1}=\frac{p\left(2\sqrt{p}+1\right)}{\left(2\sqrt{p}-1\right)\left(2\sqrt{b}+1\right)}=\frac{p\left(2\sqrt{b}+1\right)}{4p-1}\)
Bài 50 (trang 30 SGK Toán 9 Tập 1)
Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa
$\dfrac{5}{\sqrt{10}}$; $\dfrac{5}{2 \sqrt{5}}$ ; $\dfrac{1}{3 \sqrt{20}}$ ; $\dfrac{2 \sqrt{2}+2}{5 \sqrt{2}}$ ;$\dfrac{y+b.\sqrt{y}}{b.\sqrt{y}}$.
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
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5.105.2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">.
525=5.525.5=552.(5.5)=552(5)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
552.5=52" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.1320=1.20320.20=203.(20.20)=203.(20)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
203.20=22.560=2560=252.30=530" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.(22+2)5.2=(22+2).252.2=22.2+2.25.(2)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
2.2+225.2=2(2+2)5.2=2+25" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.y+byby=(y+by).yby.y=yy+by.yb.(y)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
yy+b(y)2by=yy+byby" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
y(y+b)b.y=y+bb" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
.y+byby=(y)2+byby" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">
Nguồn : Bài 50 trang 30 SGK Toán 9 tập 1 - loigiaihay.com
#Ye Chi-Lien
\(\dfrac{5}{\sqrt{10}}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\cdot\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\cdot\sqrt{20}}=\dfrac{\sqrt{20}}{60}\) ;
1) thực hiện phép tính
\(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
2) trục căn thức ở mẫu : \(\dfrac{2}{\sqrt{3}-5}\)
3) khử mẫu của biểu thức lấy căn: \(\sqrt{\dfrac{2}{5}}\)
1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)
\(=5\sqrt{3}\)
2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)
\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)
\(=\dfrac{-\sqrt{3}-5}{11}\)
3) Ta có: \(\sqrt{\dfrac{2}{5}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)
\(=\dfrac{\sqrt{10}}{5}\)
Khử mẫu của biểu thức lấy căn:
\(\sqrt{\dfrac{1}{600}};\sqrt{\dfrac{11}{540}};\sqrt{\dfrac{3}{50}};\sqrt{\dfrac{5}{98}};\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}.\)
Bài 6 (trang 10 SGK Toán 9 Tập 1)
Với giá trị nào của $a$ thì mỗi căn thức sau có nghĩa:
a)$\sqrt{\dfrac{a}{3}}$; b)$\sqrt{-5a}$; c)$\sqrt{4-a}$; d)$\sqrt{3a+7}$ ?
a
căn có nghĩa
\(\Leftrightarrow\frac{a}{3}\ge0\)
\(\Leftrightarrow a\ge0\)
b
căn có nghĩa
\(\Leftrightarrow-5a\ge0\)
\(\Leftrightarrow b\le0\left(-5\le0\right)\)
c
căn có nghĩa
\(\Leftrightarrow4-a\ge0\)
\(\Leftrightarrow-a\ge0-4\)
\(\Leftrightarrow-a\ge-4\)
\(\Leftrightarrow a\le4\)
d
căn có nghĩa
\(\Leftrightarrow3a+7\ge0\)
\(\Leftrightarrow a\ge-\frac{7}{3}\)
\(\sqrt{\dfrac{a}{3}}\) xác định vs mọi x , \(\sqrt{-5A}\) XÁC ĐỊNH A=0 , \(\sqrt{4-A}\) XÁC ĐỊNH VS A= 4 ; \(\sqrt{ }\) 3A +7 XÁC ĐỊNH KHI X= -7/3
Khử mẫu của biểu thức lấy căn
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}\)
(giả thiết các biểu thức có nghĩa)
\(=\sqrt{\dfrac{b+1}{b^2}}=\left[{}\begin{matrix}\dfrac{\sqrt{b+1}}{b}\left(b>0\right)\\-\dfrac{\sqrt{b+1}}{b}\left(-1\le b< 0\right)\end{matrix}\right.\)
Bài 54 (trang 30 SGK Toán 9 Tập 1)
Rút gọn biểu thức sau (giả thiết các biểu thức chữ đều có nghĩa):
$\dfrac{2+\sqrt{2}}{1+\sqrt{2}}$ ; $\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}$ ; $\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}$ ; $\dfrac{a-\sqrt{a}}{1-\sqrt{a}}$ ; $\dfrac{p-2 \sqrt{p}}{\sqrt{p}-2}$.
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}(\sqrt{2}+1)}{1+\sqrt{2}}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}=\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}(\sqrt{a}-1)}{1-\sqrt{a}}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}(\sqrt{p}-2)}{\sqrt{p}-2}=\sqrt{p}\)