a/ ĐKXĐ: \(x^2+3x+2\ge0\)
\(\Leftrightarrow3-2\sqrt{x^2+3x+2}=1-2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\sqrt{x^2+3x+2}=\sqrt{x^2-x+1}+1\)
\(\Leftrightarrow x^2+3x+2=x^2-x+1+1+2\sqrt{x^2-x+1}\)
\(\Leftrightarrow2x=\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=x^2-x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3x^2+x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{-1+\sqrt{13}}{6}\\x=\frac{-1-\sqrt{13}}{6}\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(3x^2-7x+2\ge0\)
\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\) (1)
\(\Rightarrow3x^2-5x+7=9+3x^2-7x+2-6\sqrt{3x^2-7x+2}\)
\(\Rightarrow2-x=3\sqrt{3x^2-7x+2}\) (\(x\le2\))
\(\Rightarrow\left(2-x\right)^2=9\left(3x^2-7x+2\right)\)
\(\Rightarrow x^2-4x+4=27x^2-63x+18\)
\(\Rightarrow26x^2-59x+14=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)
Do bước biến đổi thứ 2 ko phải phép tương đương nên cần thay 2 nghiệm vào (1) để kiểm tra lại, bạn tự thay nhé
c/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1\\x\le-5\end{matrix}\right.\)
\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+2\right)\left(x+1\right)\left(x+5\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)
d/ ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\1\le x\le5\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x^2-1}+\sqrt{5-x}=\sqrt{x^2+3}\)
\(\Leftrightarrow x^2-x+4+2\sqrt{\left(x^2-1\right)\left(5-x\right)}=x^2+3\)
\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(5-x\right)}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\4\left(x^2-1\right)\left(5-x\right)=\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left(x-1\right)\left[4\left(x+1\right)\left(5-x\right)-x+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-4x^2+15x+21=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{15+\sqrt{561}}{8}\\x=\frac{15-\sqrt{561}}{8}\left(l\right)\end{matrix}\right.\)
e/
\(\Leftrightarrow\left(x-1\right)\sqrt{1+x\sqrt{x^2+4}}-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\sqrt{1+x\sqrt{x^2+4}}-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\sqrt{1+x\sqrt{x^2+4}}=x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1+x\sqrt{x^2+4}=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow1+x\sqrt{x^2+4}=x^2+2x+1\)
\(\Leftrightarrow x\left(\sqrt{x^2+4}-x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+4}=x+2\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^2+4=\left(x+2\right)^2\) (vẫn sử dụng điều kiện \(x\ge-1\) của (1))
\(\Leftrightarrow x^2+4=x^2+4x+4\)
\(\Rightarrow x=0\)
