Rút gọn biểu thức :a) \(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\) với 0 \(\le\)x \(\le\)yb) \(\frac{3-\sqrt{x}}{x-9}\)(với x \(\ge\)0 , x \(\ne\)9)c) \(\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}\)( với x\(\ge\)0 , x\(\ne\)9)d) \(6-2x-\sqrt{9-6x+x^2}\)(với x...
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Rút gọn biểu thức :
a) \(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\) với 0 \(\le\)x \(\le\)y
b) \(\frac{3-\sqrt{x}}{x-9}\)(với x \(\ge\)0 , x \(\ne\)9)
c) \(\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}\)( với x\(\ge\)0 , x\(\ne\)9)
d) \(6-2x-\sqrt{9-6x+x^2}\)(với x <3)
\(a,\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\left|\sqrt{x}-\sqrt{y}\right|\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=y-x\)
\(b,\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
\(c,\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(d,6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-3+x=3-x\)
\(a,\)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(=|\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)|\)
\(=|\sqrt{x}^2-\sqrt{y}^2|\)
\(=|x-y|\)
Vì \(x\le y\)\(\Rightarrow x-y\ge0\)
\(\Rightarrow|x-y|=x-y\)