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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 ) \(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)
\(=\sqrt{\dfrac{2\left(5-\sqrt{21}\right)}{2}}-\sqrt{\dfrac{2\left(5+\sqrt{21}\right)}{2}}\)
\(=\sqrt{\dfrac{10-2\sqrt{21}}{2}}-\sqrt{\dfrac{10+2\sqrt{21}}{2}}\)
\(=\dfrac{\sqrt{7-2\sqrt{21}+3}}{\sqrt{2}}-\dfrac{\sqrt{7+2\sqrt{21}+3}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
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a. =\(\frac{x\sqrt{xy}+y\sqrt{x^2}-x\sqrt{y^2}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}\)
=\(\frac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}\)=\(\frac{\sqrt{xy}\left(x-y\right)}{\sqrt{xy}}\)=\(x-y\)
b. =\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x-1}}\)=\(x+\sqrt{x}+1\)
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
1/ Thực hiện phép tính
a) √9−2√20+√12−2√35
\(=\sqrt{\left(\sqrt{5}-\sqrt{4}\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}\)
\(=\sqrt{5}-\sqrt{4}+\sqrt{7}-\sqrt{5}=\sqrt{7}-\sqrt{4}=\sqrt{7}-2\)