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a) Ta có: \(2\sqrt{8}-3\sqrt{18}+\sqrt{32}\)
\(=4\sqrt{2}-6\sqrt{2}+4\sqrt{2}\)
\(=2\sqrt{2}\)
b) Ta có: \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1+\sqrt{2}+1\)
\(=2\sqrt{2}\)
c) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
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a: =>\(\left(\dfrac{2x+1}{9}+1\right)+\left(\dfrac{2x+2}{8}+1\right)+...+\left(\dfrac{2x+9}{1}+1\right)=0\)
=>2x+10=0
=>x=-5
b: \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+...+\left(\dfrac{x-2014}{2}-1\right)+\left(x-2016\right)=0\)
=>x-2016=0
=>x=2016
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a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(=\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x^2-x-2\right)}{x^2}\)
\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:
\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Ta có: \(x\left(x-4\right)-x^2+8=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2