đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{cases}\left(a,b\ge0\right)}\)

\(\Rightarrow\hept{\begin{cases}x^2-x+1=b^2\\\sqrt{x^3+1}=\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=ab\end{cases}}\)

PT tương đương với : 

\(x^2-x+1+2\sqrt{\left(x+1\right)\left(x^2-x+1\right)}-1=2\sqrt{x+1}\)

\(\Leftrightarrow b^2+2ab-1=2a\Leftrightarrow b^2+2ab+a^2=a^2+2a+1\)

\(\Leftrightarrow\left(a+b\right)^2=\left(a+1\right)^2\Leftrightarrow\orbr{\begin{cases}a+b=a+1\\a+b=-\left(a+1\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}b=1\\loai\left(VT\ge0;VP< 0\right)\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-x+1}=1\Leftrightarrow x^2-x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}\left(tm\right)}\)

Vậy ...

\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)

\(=\left(\frac{-x+1}{\sqrt{2}}\right)^2.\frac{-2x-2\sqrt{x}}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}\)

\(=\frac{-2x-2\sqrt{x}}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{\left(-x+1\right)^2}{2}\)

\(=\frac{\left(-2x-2\sqrt{x}\right)\left(1-x\right)^2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right).2}\)

\(=\frac{\left(-2x-2\sqrt{x}\right)\left(-x+1\right)}{2\left(x+2\sqrt{x}+1\right)}\)

\(=\frac{2\left(x+\sqrt{x}\right)\left(-x+1\right)}{2\left(x+2\sqrt{x}+1\right)}\)

\(=\frac{\left(x+\sqrt{x}\right)\left(-x+1\right)}{\left(x+2\sqrt{x}+1\right)}\)

\(=\frac{\left(x+\sqrt{x}\right)\left(-x+1\right)}{x+2\sqrt{x}+1}\)

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