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a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)

b: \(B=5x^2-7x\sqrt{y}+2y\)

\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)

\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)

\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)

Bài 3:

a: \(A=\frac{1}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{2\sqrt{x}}{4-x}\)

\(=\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{\sqrt{x}-2-\sqrt{x}-2-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=-\frac{2}{\sqrt{x}-2}\)

b: Thay x=3 vào A, ta được: \(A=-\frac{2}{\sqrt3-2}=\frac{2}{2-\sqrt3}=2\left(2+\sqrt3\right)=4+2\sqrt3\)

Bài 2:

a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)

\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)

b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)

Bài 1:

a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)

\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)

b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)

\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)

\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)

3 tháng 11 2015

\(M=7\sqrt{x-1}-\sqrt{x^3-x^2}+x-1\)

\(=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}-\left(\sqrt{x-1}\right)^2\)

\(=\sqrt{x-1}\left(7-\sqrt{x^2}-\sqrt{x-1}\right)\)

\(=\sqrt{x-1}\)(7 - /x/ - \(\sqrt{x-1}\) )

20 tháng 9 2020

x - 3 = ( √x )2 - ( √3 )2 = ( √x - √3 )( √x + √3 ) < với x > 0 >

20 tháng 9 2020

            Bài làm :

Ta có :

\(x-3=\left(\sqrt{x}\right)^2-\left(\sqrt{3}\right)^2=\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)\)