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A=2x-|2x+1|
TH1: x>=-1/2
A=2x-2x-1=-1
TH2: x<-1/2
A=2x+2x+1=4x+1
\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)
\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)
a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)
\(=\dfrac{1}{x}\)
b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)
\(=\dfrac{3x+6}{x-2}\)
\(1,\\ a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x+1}=\dfrac{2}{3}\Leftrightarrow2x+1=\dfrac{4}{9}\Leftrightarrow x=-\dfrac{5}{18}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=2\Leftrightarrow\left[{}\begin{matrix}x-3=2\\3-x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\ 2,\\ a,=\left|5-x\right|=x-5\\ b,=\sqrt{4a\cdot44a}=\sqrt{176a^2}=4\left|a\right|\sqrt{11}=4a\sqrt{11}\\ c,=\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)
a) \(\Leftrightarrow\)\(\sqrt{4x+8}\) + \(2\sqrt{x+2}\) \(-\sqrt{9x}\)\(-\)18 = 1 (Đkxd: x \(\ge\)0)
\(\Leftrightarrow\)\(2\sqrt{x+2}+2\sqrt{x+2}-\sqrt{9x}=19\)
\(\Leftrightarrow\)\(4\sqrt{x+2}=19+\sqrt{9x}\)
\(\Leftrightarrow16x+32=361+2\times19\sqrt{9x}+9x\)
\(\Leftrightarrow7x=329+144\sqrt{x}\)
\(\Leftrightarrow49x-114\times7\sqrt{x}+3249=5552\)
\(\Leftrightarrow\left(7\sqrt{x}-57\right)^2=5552\)
\(\Leftrightarrow7\sqrt{x}-57=\pm4\sqrt{347}\)
Từ đó bạn tự tìm ra x nhé . Mình hơi bận nên không giải hết được
\(A=\dfrac{2x+1}{4x-2}+\dfrac{1-2x}{4x+2}-\dfrac{2}{1-4x^2}\)
\(=\dfrac{2x+1}{2\left(2x-1\right)}+\dfrac{1-2x}{2\left(2x+1\right)}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)\(=\dfrac{\left(2x+1\right)^2+\left(1-2x\right)^2+4}{2\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{4x^2+4x+1+1-4x+4x^2+4}{2\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{8x^2+6}{2\left(4x^2-1\right)}=\dfrac{8x^2-2+8}{8x^2-2}=1+\dfrac{8}{2\left(4x^2-1\right)}\)\(=1+\dfrac{4}{4x^2-1}\)