rút gọn rồi tính
\(A=\sqrt{\frac{1}{x^2-4x+4}}-\frac{4}{x^2-4}\) \(\left(x>2\right);x=3\)
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26 tháng 8 2020
a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
30 tháng 7 2016
b) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}\)
\(=4x-\sqrt{8}+\frac{x\left(x+2\right)}{x+2}\)
\(=4x-\sqrt{8}+x\)
\(=5x-\sqrt{8}\)
Với \(x=-\sqrt{2}\) ta có:
\(5x-\sqrt{8}=5\cdot\left(-\sqrt{2}\right)-\sqrt{4\cdot2}=-5\sqrt{2}-2\sqrt{2}=-7\sqrt{2}\)
Rút gọn:
\(A=\sqrt{\frac{1}{x^2-4x+4}}+\frac{-4}{x^2-2^2}\)
\(=\sqrt{\frac{1}{\left(x-2\right)^2}}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{1}{x-2}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2-4}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x+2}\)
Thay x=3 vào A ta được \(\frac{1}{3+2}=\frac{1}{5}\)