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ĐKXĐ ; \(x\ne\pm1\)
Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}+\dfrac{-x^2-3}{x^2-1}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-x^2-3=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3=0\)
\(\Leftrightarrow-x^2+4x-3=0\)
\(\Leftrightarrow-x^2+3x+x-3=0\)
\(\Leftrightarrow-x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=1\left(L\right)\end{matrix}\right.\)
=> X = 3
Vậy ..
a) Có \(x+1< x+2\)
\(\Rightarrow\sqrt{x+1}< \sqrt{x+2}\)
\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+2}}< 1\)
b) Vì \(\sqrt{x+1}< \sqrt{x+2}\)
\(\Rightarrow\sqrt{x+1}.\sqrt{x+1}.\sqrt{x+2}< \sqrt{x+2}.\sqrt{x+1}.\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{x+1}^2.\sqrt{x+2}< \sqrt{x+2}^2.\sqrt{x+1}\)
\(\Rightarrow\frac{\sqrt{x+1}^2}{\sqrt{x+2}^2}< \frac{\sqrt{x+1}}{\sqrt{x+2}}\)
hay \(\frac{\sqrt{x+1}}{\sqrt{x+2}}>\frac{\sqrt{x+1}^2}{\sqrt{x+2}^2}\)
Ta có : \(\frac{x-1}{-1}+\frac{1}{x+1}=\frac{2x-1}{x^2+x}\)
\(\Rightarrow\) \(-x+1=\frac{2x-1}{x\left(x+1\right)}-\frac{1}{x+1}\) \(=\frac{1}{x+1}.\left(\frac{2x-1-x}{x}\right)=\frac{1}{x+1}.\frac{x-1}{x}\)
\(\Rightarrow\frac{1}{x+1}.\frac{x-1}{x}+x-1=0\)
\(\Rightarrow\frac{x-1}{x\left(x+1\right)}+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(\frac{1}{x\left(x+1\right)}+1\right)=0\)
\(\Rightarrow x-1=0\)hoặc \(\frac{1}{x\left(x+1\right)}+1=0\)
\(\Rightarrow x=1\) hoặc \(\frac{1}{x\left(x+1\right)}=-1\)
\(\Rightarrow x=1\)hoặc \(x^2+x=-1\)
\(\Rightarrow x=1\)hoặc \(x^2+1+x=0\)
\(\Rightarrow x=1\)hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\)
\(\Rightarrow x=1\)hoặc \(\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\)(vô lí)
Vậy : x = 1 thì thỏa mãn điều kiện bài toán.