\(\sqrt{\frac{x+1}{x^2-x+1}}+\sqrt{\frac{x^2-x+1}{x+1}}=2\)
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24 tháng 7 2017
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
\(\sqrt{\frac{x+1}{x^2-x+1}}+\sqrt{\frac{x^2-x+1}{x+1}}=2\)
\(\Leftrightarrow\left(\sqrt{\frac{x+1}{x^2-x+1}}+\sqrt{\frac{x^2-x+1}{x+1}}\right)^2=2^2\)
\(\Leftrightarrow\frac{x+1}{x^2-x+1}+2.\sqrt{\frac{x+1}{x^2-x+1}}.\sqrt{\frac{x^2-x+1}{x+1}}+\frac{x^2-x+1}{x+1}=4\)
\(\Leftrightarrow\frac{x+1}{x^2-x+1}+\frac{x^2-x+1}{x+1}=4-2\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x^2-x+1\right)\left(x+1\right)}+\frac{\left(x^2-x+1\right)^2}{\left(x+1\right)\left(x^2-x+1\right)}=2\)
\(\Leftrightarrow\frac{\left(x+1\right)^2+\left(x^2+1-x\right)^2}{x^3+1}=2\)
\(\Leftrightarrow\frac{x^2+2x+1+x^4+1+x^2+2x^2-2x-2x^3}{x^3+1}=2\)
\(\Leftrightarrow x^2+2x+1+x^4+1+x^2+2x^2-2x-2x^3=2\left(x^3+1\right)\)
\(\Leftrightarrow4x^2+2+x^4-2x^3=2x^3+2\)
\(\Leftrightarrow x^4-2x^3-2x^3+4x^2=2-2\)
\(\Leftrightarrow x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-4x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-2\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bạn tự tìm ĐKXĐ rồi so sánh kết quả nhé!