(3-x)\(\sqrt{x+3}+\left(x+4\right)\sqrt{2-x}=2\sqrt{\left(x+3\right)\left(2-x\right)+5}\)
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Những câu hỏi liên quan
a. \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
b. \(3\left(2+\sqrt{x}\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
c. \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
d. \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
\(=x-3\sqrt{x}-5\sqrt{x}-15\)
\(=x-8\sqrt{x}-15\)
b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
\(=3\sqrt{x}+6+2\sqrt{x}-x+6-3\sqrt{x}\)
\(=-x+2\sqrt{x}+12\)
c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
\(=x-9-5\sqrt{x}+5\)
\(=x-5\sqrt{x}-4\)
d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3\sqrt{x}-6-x+1\)
\(=-x+3\sqrt{x}-5\)
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Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
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Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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b, MA-Bfrac{sqrt{x}+2}{sqrt{x}+3}-left(frac{5}{x+sqrt{x}-6}+frac{1}{sqrt{x}-2}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}-frac{1}{sqrt{x}-2}frac{left(sqrt{x}+2right)left(sqrt{x}-2right)}{x+sqrt{x}-6}-frac{5}{x+sqrt{x}-6}-frac{1left(sqrt{x}+3right)}{x+sqrt{x}-6}frac{x-4-5-sqrt{x}-3}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-4sqrt{x}+3sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{left(sqrt{x}-4right)left(sqrt{x...
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b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
bạn trung học hay tiểu học vậy
Rút gọn :
dfrac{sqrt{x+sqrt{4left(x-1right)}}-sqrt{x-sqrt{4left(x-1right)}}}{sqrt{x^2-4left(x-1right)}}.left(sqrt{x-1}-dfrac{1}{sqrt{x-1}}right)
b)left(sqrt{2}+1right)left(sqrt{3}+1right)left(sqrt{6}+1right)left(5-2sqrt{2}-sqrt{3}right)
c)left(sqrt{5}+1right)left(sqrt{7}+1right)left(sqrt{35}+1right)left(34-4sqrt{7}-6sqrt{5}right)
d) left(sqrt{7}+1right)left(2sqrt{2}-1right)left(2sqrt{14}-1right)left(55+12sqrt{2}-7sqrt{7}right)
e)left(3sqrt{2}+1right)left(2sqrt{3}+1right)left(6sqrt{6}+1...
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Rút gọn :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
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tínhsqrt{left(2-sqrt{5}right)^2}+sqrt{left(2sqrt{2}-sqrt{5}right)^2}sqrt{left(sqrt{7}-2sqrt{2}right)^2}+sqrt{left(3-2sqrt{2}right)^2}sqrt{left(x-3right)^2}left(x3right)sqrt{left(1-xright)^2}left(x1right)sqrt{9a^4}sqrt{100a^2}left(a 0right)
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tính
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(\sqrt{\left(x-3\right)^2}\left(x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\left(x>1\right)\)
\(\sqrt{9a^4}\)
\(\sqrt{100a^2}\left(a< 0\right)\)
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)
\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)
\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)
\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)
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Lời giải:
a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$
$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$
$=3-\sqrt{7}$
c.
$=|x-3|=x-3$
d.
$=|1-x|=x-1$
$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.
$=\sqrt{(10a)^2}=|10a|=-10a$
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giải pt
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1
Giải phương trình
\(\frac{3\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{5\left(x-1\right)\left(x+\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)
Giải các phương trình sau:a 2sqrt[3]{left(x+2right)^2}-sqrt[3]{left(x-2right)^2}sqrt[3]{x^2-4}b sqrt[3]{left(65+xright)^2}+4sqrt[3]{left(65-xright)^2}5sqrt[3]{65^2-x^2}c sqrt[3]{x+1}+sqrt[3]{x+2}1+sqrt[3]{x^2+3x+2}d sqrt[3]{x-2}+sqrt[3]{x+3}sqrt[3]{2x+1}e sqrt[3]{2x-1}+sqrt[3]{x-1}sqrt[3]{3x+1}
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Giải các phương trình sau:
a \(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
b \(\sqrt[3]{\left(65+x\right)^2}+4\sqrt[3]{\left(65-x\right)^2}=5\sqrt[3]{65^2-x^2}\)
c \(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
d \(\sqrt[3]{x-2}+\sqrt[3]{x+3}=\sqrt[3]{2x+1}\)
e \(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}\)
a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
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b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
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c.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x+1}=b\end{matrix}\right.\)
\(\Rightarrow a+b=1+ab\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=1\\b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=1\\x+1=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
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