\(\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}\)(0<x<1)
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5 tháng 8 2019

\(E=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}\)

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

\(E=\frac{2}{x-1}\cdot\frac{-\left(x-1\right)}{2x}\)

\(E=\frac{-1}{x}\)

_________

\(G=\frac{x-16}{\sqrt{x-7}-3}\)

\(G=\frac{\left(\sqrt{x-7}-3\right)\left(\sqrt{x-7}+3\right)}{\sqrt{x-7}-3}\)

\(G=\sqrt{x-7}+3\)

20 tháng 5 2019

a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)

20 tháng 5 2019

\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\) 

        \(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)  

\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\)  \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)

7 tháng 8 2017

1. \(N=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)

\(N=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(N=\left(\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(N=\left(\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(N=\left(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right).\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

\(N=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

\(N=\frac{4x}{x-3}\)

Vậy \(N=\frac{4x}{x-3}\)với \(x>0,x\ne4,x\ne9\)

2.Với \(x>0,x\ne4,x\ne9\)

Ta có \(N< 0\)\(\Leftrightarrow\frac{4x}{x-3}< 0\)\(\Leftrightarrow x-3< 0\)(Vì \(x>0\Leftrightarrow4x>0\)\(với\forall x\))\(\Leftrightarrow x< 3\)

Vậy ..........

3. Với \(x>0,x\ne4,x\ne9\)

Ta có \(\left|N\right|=1\Leftrightarrow\left|\frac{4x}{x-3}\right|=1\Leftrightarrow\orbr{\begin{cases}\frac{4x}{x-3}=1\\\frac{4x}{x-3}=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}4x=3-x\\4x=x-3\end{cases}}\)\(\orbr{\begin{cases}x=\frac{3}{5} \left(N\right)\\x=-1\left(N\right)\end{cases}}\)

Vậy ...............