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\(=\frac{2\sqrt{x}}{x-3}.\frac{\sqrt{\left(x-3\right)^2}}{\sqrt{x}}=\frac{2\left(x-3\right)}{x-3}=-2\)
a) \(4x-\sqrt{x^2-4x+4}=4x-\sqrt{\left(x-2\right)^2}=4x-\left(x-2\right)=3x+2\)
b) \(3x+\sqrt{9+6x+x^2}=3x+\sqrt{\left(x+3\right)^2}=3x-\left(x+3\right)=2x-3\)
c) \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
d) \(\frac{\sqrt{x^2+4x+4}}{x+2}=\frac{\sqrt{\left(x+2\right)^2}}{x+2}=\frac{\left|x+2\right|}{x+2}\)( 1 )
với x < -2 thì : \(\left(1\right)\Leftrightarrow\frac{-\left(x+2\right)}{x+2}=-1\)
với x > -2 thì : \(\left(1\right)\Leftrightarrow\frac{\left(x+2\right)}{x+2}=1\)
Ta có : \(A=\sqrt{\left(x+3\right)^2}-\frac{\sqrt{x^2-6x+9}}{x+3}\)
\(\Rightarrow A=\left(x+3\right)-\frac{\left(x-3\right)^2}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{\left(x+3\right)^2}{\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{\left(x+3+x-3\right)\left(x+3-x+3\right)}{\left(x+3\right)}\)
\(\Rightarrow A=\frac{18x}{\left(x+3\right)}\)
Mình quên ghi điều kiện mất là x<-3 nhé
Mấy bạn làm lại giúp mình
Mình cảm ơn :D
\(a,\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\left|\sqrt{x}-\sqrt{y}\right|\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=y-x\)
\(b,\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
\(c,\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(d,6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-3+x=3-x\)
\(a,\)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(=|\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)|\)
\(=|\sqrt{x}^2-\sqrt{y}^2|\)
\(=|x-y|\)
Vì \(x\le y\)\(\Rightarrow x-y\ge0\)
\(\Rightarrow|x-y|=x-y\)
\(\frac{2}{x-3}\sqrt{\frac{x^2-6x+9}{4y^4}}=\frac{2}{x-3}.\frac{\sqrt{x^2-6x+9}}{\sqrt{4y^4}}=\frac{2}{x-3}.\frac{\sqrt{\left(x-3\right)^2}}{\sqrt{\left(2y^2\right)^2}}\)
\(=\frac{2}{x-3}.\frac{x-3}{2y^2}=\frac{1}{y^2}\)
\(\frac{2}{x-3}\sqrt{\frac{x^2-6x+9}{4y^4}}\)
\(=\frac{2}{x-3}\sqrt{\frac{\left(x-3\right)^2}{\left(2y^2\right)^2}}\)
\(=\frac{2}{x-3}.\left|\frac{x-3}{2y^2}\right|\)
\(=\frac{2}{x-3}.\frac{3-x}{2y^2}\)( vi \(x< 3;y\ne0\))
\(=\frac{-2}{2y^2}\)
\(=\frac{-1}{y^2}\)
a)\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+x-3\)
\(=2x\)
b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\sqrt{\left(x+2\right)^2}-x\)
\(=x+2-x\)
=2
c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)
\(=\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(=\sqrt{x-1}\)
\(\sqrt{\frac{x^2-6x+9}{x-3}}=\sqrt{\frac{\left(x-3\right)^2}{x-3}}=\sqrt{x-3}\)