\(=\left(\right)\dfrac{2+5}{3}-\sqrt{3}\) ).2\(\sqrt{3}\)
=\(\dfrac{14}{\sqrt{3}}-6\)
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Bài 3: Liên hệ giữa phép nhân và phép khai phương
Các câu hỏi tương tự
bài 1: tính
a) sqrt{1,2cdot27} b) sqrt{55cdot77cdot35}
c) (sqrt{3}-sqrt{2} )^2 d) (3sqrt{2}-1)*(3sqrt{2}+1)
e) (sqrt{6}+7) (sqrt{3}-sqrt{2}) i) sqrt{dfrac{1}{8}}cdotsqrt{2}cdotsqrt{125}cdotsqrt{dfrac{1}{5}}
h) sqrt{sqrt{2}-1}cdotsqrt{sqrt{2}}+1
bài 2: tính
a) sqrt{9}-sqrt{17}cdotsqrt{9}+sqrt{17}
b) 2sqrt{2}left(sqrt{3}-2right)+left(1+2sqrt{2}right)^2-2sqrt{6}
c) dfrac{sqrt{6}+sqrt{14}}{2sqrt{3}+sqrt{28}} d) dfrac{sqrt{10}+sqrt{15}}{sqrt{8}+sqrt{...
Đọc tiếp
bài 1: tính
a) \(\sqrt{1,2\cdot27}\) b) \(\sqrt{55\cdot77\cdot35}\)
c) (\(\sqrt{3}-\sqrt{2}\) )\(^2\) d) (3\(\sqrt{2}-1\))*(3\(\sqrt{2}+1\))
e) (\(\sqrt{6}+7\)) (\(\sqrt{3}-\sqrt{2}\)) i) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\)
h) \(\sqrt{\sqrt{2}-1}\cdot\sqrt{\sqrt{2}}+1\)
bài 2: tính
a) \(\sqrt{9}-\sqrt{17}\cdot\sqrt{9}+\sqrt{17}\)
b) 2\(\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
c) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) d) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\) f) \(\dfrac{x+\sqrt{xy}}{9+\sqrt{xy}}\) (xy>0)
Rút gọn các biểu thức sau :a,dfrac{sqrt{6}+sqrt{10}}{sqrt{21}+sqrt{35}}b,dfrac{sqrt{405}+3sqrt{27}}{3sqrt{3}+sqrt{45}}c,dfrac{sqrt{2}+sqrt{3}+sqrt{4}-sqrt{6}-sqrt{9}-sqrt{12}}{sqrt{2}+sqrt{3}+sqrt{4}}d, Ddfrac{2}{x^2-y^2}cdotsqrt{dfrac{9left(x^2+2xy+y^2right)}{4}} left(vớixne y,xne-yright)
Đọc tiếp
Rút gọn các biểu thức sau :
a,\(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b,\(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c,\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d, D=\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\) \(\left(vớix\ne y,x\ne-y\right)\)
B2 : Tính : a, left(sqrt{x}-3right).left(sqrt{x}+2right)b, left(sqrt{x}-sqrt{y}right).left(sqrt{x}+sqrt{y}right)c, left(sqrt{dfrac{25}{3}}-sqrt{dfrac{49}{3}}+sqrt{3}right).sqrt{3}d,left(1+sqrt{3}-sqrt{5}right).left(1+sqrt{3}+sqrt{5}right)
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B2 : Tính :
a, \(\left(\sqrt{x}-3\right)\)\(.\left(\sqrt{x}+2\right)\)
b, \(\left(\sqrt{x}-\sqrt{y}\right).\)\(\left(\sqrt{x}+\sqrt{y}\right)\)
c, \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right)\)\(.\sqrt{3}\)
d,\(\left(1+\sqrt{3}-\sqrt{5}\right)\)\(.\left(1+\sqrt{3}+\sqrt{5}\right)\)
Rút gọn:
a)\(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b)\(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c)\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d)\(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}\)
Tính
a) \(\sqrt{13-4\sqrt{2}}\)
b) \(2\sqrt{40\sqrt{12}}-2\sqrt{75}-3\sqrt{5\sqrt{48}}\)
c) \(\sqrt{1+\dfrac{1}{1^2}+\dfrac{2}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{2}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{2}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{2}{100^2}}\)
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
A)sqrt{2-sqrt{3}}.left(sqrt{6}+sqrt{2}right)
B)left(sqrt{2}+1^{ }right)^3-left(sqrt{2}-1right)^3 C)dfrac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}-dfrac{sqrt{5}+sqrt{27}}{sqrt{30}+sqrt{162}} D)sqrt{dfrac{2-sqrt{3}}{2+sqrt{3}}}+sqrt{dfrac{2+sqrt{3}}{2-sqrt{3}}} E)dfrac{sqrt{3-sqrt{5}}.left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}} F)dfrac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+dfrac{1}{sqrt{2}-sqrt{2-sqrt{3}}}
Đọc tiếp
A)\(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
B)\(\left(\sqrt{2}+1^{ }\right)^3-\left(\sqrt{2}-1\right)^3\) C)\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) D)\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\) E)\(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\) F)\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
H24
30 tháng 5 2021 lúc 18:18
Cho x= \(\dfrac{\sqrt{13-4\sqrt{3}}}{2}.Tính\) A = \(\dfrac{\sqrt{x}+2}{2\sqrt{x}-3}\)
Thực hiện phép tính:
a) \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\sqrt{\dfrac{1}{ab}}}\right)\cdot\sqrt{ab}\)
b) \(\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)