c.
ĐLXĐ: \(x\ge-\dfrac{1}{3}\)
\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)
Đặt \(\sqrt{3x+1}=t\ge0\)
\(\Rightarrow-t^2+t+4x^2-10x+6=0\)
\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-\dfrac{61}{12}\)
\(\Leftrightarrow36x^2+12x-58-2\sqrt{12x+61}=0\)
\(\Leftrightarrow\left(36x^2+24x+4\right)-\left(12x+61+2\sqrt{12x+61}+1\right)=0\)
\(\Leftrightarrow\left(6x+2\right)^2-\left(\sqrt{12x+61}+1\right)^2=0\)
\(\Leftrightarrow\left(6x+1-\sqrt{12x+61}\right)\left(6x+3+\sqrt{12x+61}\right)=0\)
\(\Leftrightarrow...\) tương tự câu a
a.
ĐKXĐ: \(x\ge-\dfrac{5}{4}\)
\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)
\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)
\(\Leftrightarrow\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)
\(\Leftrightarrow\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)
\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
