a/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2+5x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\) pt trở thành:
\(a-2b=a^2-4b^2\)
\(\Leftrightarrow a-2b=\left(a-2b\right)\left(a+2b\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}a-2b=0\\a+2b=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\left(1\right)\\a+2b=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x^2+5x+1}=2\sqrt{x^2-x+1}\)
\(\Leftrightarrow4x^2+5x+1=4\left(x^2-x+1\right)\)
\(\Leftrightarrow...\)
Kết hợp (2) với pt ban đầu ta được: \(\left\{{}\begin{matrix}a-2b=9x-3\\a+2b=1\end{matrix}\right.\)
\(\Rightarrow2a=9x-2\)
\(\Rightarrow2\sqrt{4x^2+5x+1}=9x-2\) (\(x\ge\frac{2}{9}\))
\(\Leftrightarrow4\left(4x^2+5x+1\right)=\left(9x-2\right)^2\)
\(\Leftrightarrow...\)
b/ ĐKXĐ:
\(\left(x+y\right)^2-1+\sqrt{2x+2y}-\sqrt{x+y+1}=0\)
\(\Leftrightarrow\left(x+y+1\right)\left(x+y-1\right)+\frac{x+y-1}{\sqrt{2x+2y}+\sqrt{x+y+1}}=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x+y+1+\frac{1}{\sqrt{2x+2y}+\sqrt{x+y+1}}\right)=0\)
\(\Leftrightarrow x+y-1=0\)
\(\Leftrightarrow y=1-x\)
Thay xuống pt dưới:
\(x^2-x\left(1-x\right)=3\)
\(\Leftrightarrow2x^2-x-3=0\)