a,\(\sqrt{\sqrt{\left(\sqrt{3-1}\right)^4}}\)
b,\(\sqrt{\sqrt{x^4}}\)vs \(x\ge0\)
M.N HELP T VS~~~ Ô - MAI - GÓT
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10 tháng 9 2017
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11 tháng 11 2018
\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)
\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)
\(=x-2\sqrt{x}+4\)
=.= hok tốt!!
21 tháng 10 2018
Bài 1 : Thực hiện phép tính :
a ) \(\sqrt{9-2\sqrt{20}}+\sqrt{12-2\sqrt{35}}\)
\(=\sqrt{5-2\sqrt{20}+4}+\sqrt{7-2\sqrt{35}+5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{5}\right)}^2\)
\(=\sqrt{5}-2+\sqrt{7}-\sqrt{5}\)
\(=\sqrt{7}-2\)
b ) \(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)
\(=\sqrt{\dfrac{2\left(5-\sqrt{21}\right)}{2}}-\sqrt{\dfrac{2\left(5+\sqrt{21}\right)}{2}}\)
\(=\sqrt{\dfrac{10-2\sqrt{21}}{2}}-\sqrt{\dfrac{10+2\sqrt{21}}{2}}\)
\(=\dfrac{\sqrt{7-2\sqrt{21}+3}}{\sqrt{2}}-\dfrac{\sqrt{7+2\sqrt{21}+3}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
11 tháng 7 2017
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
14 tháng 8 2019
1) \(B=\sqrt{x-1+2\sqrt[3]{x\sqrt{x}+3x+3\sqrt{x}+1}}\)
\(B=\sqrt{x-1+2\sqrt[3]{\sqrt{x^3}+3x+3\sqrt{x}+1}}\)
\(B=\sqrt{x-1+2\sqrt[3]{\left(\sqrt{x}+1\right)^3}}\)
\(B=\sqrt{x-1+2\left(\sqrt{x}+1\right)}\)
\(B=\sqrt{x-1+2\sqrt{x}+2}\)
\(B=\sqrt{\left(\sqrt{x}+1\right)^2}\)
\(B=\sqrt{x}+1\)
\(B=\sqrt{5}+1\)
2) Sửa đề :
\(C=\sqrt{2x-1+2\sqrt{x^2-x}}+\sqrt{2x-1-2\sqrt{x^2-x}}\)
\(C=\sqrt{x+2\sqrt{x\left(x-1\right)}+x-1}+\sqrt{x-2\sqrt{x\left(x-1\right)}+x-1}\)
\(C=\sqrt{\left(\sqrt{x}+\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x}-\sqrt{x-1}\right)^2}\)
\(C=\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\)
\(C=2\sqrt{x}\)
\(C=2\cdot\sqrt{4}=4\)
LD
19 tháng 8 2017
a) Q=\(\left(\dfrac{2x+1}{\sqrt{x}^3-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)
=\(\left(\dfrac{2x+1-x+\sqrt{x}}{\sqrt{x}^3-1}\right)\left(\dfrac{1+\sqrt{x}^3-\sqrt{x}-x}{1+\sqrt{x}}\right)\)
=\(\dfrac{\sqrt{x}+x+1}{\sqrt{x}^3-1}.\left(-2\sqrt{x}+1\right)\)
=\(\dfrac{\left(-2\sqrt{x}+1\right)\left(\sqrt{x}+x+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)=\(\dfrac{\left(-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
b) ta có : Q=3 => \(\dfrac{-2\sqrt{x}+1}{\sqrt{x}-1}=3=>-2\sqrt{x}+1=3\sqrt{x}-3\)
=>x=16/25=0,64
vậy x=0,64 khi Q=3
a)\(\sqrt{\sqrt{\left(\sqrt{3-1}\right)^4}}\)\(=\sqrt{\left(\sqrt{3-1}\right)^2}\)
\(=\sqrt{3-1}=\sqrt{2}\)
b)\(\sqrt{\sqrt{x^4}}=\sqrt{x^2}=x\)