\(A=\sqrt{4x+1}-\sqrt{3x-2}=\frac{x+3}{2}\)
đkxđ \(\hept{\begin{cases}x\ge-\frac{1}{4}\\x\ge\frac{2}{3}\end{cases}}\)
đặt t=x+3 phương trình trở thành
\(A=\sqrt{4\left[x+3\right]-11}-\sqrt{3\left[x+3\right]-11}=\frac{x+3}{2}\)
\(A=\sqrt{4t-11}-\sqrt{3t-11}=\frac{t}{2}\)
\(\Leftrightarrow4t-11=\frac{t^2}{4}+3t-11+t\sqrt{3t-11}\)
\(\Leftrightarrow t^2-\frac{t^2}{4}=t\sqrt{3t-11}\)
\(\Leftrightarrow\frac{t\left[4-t\right]}{4}=t\sqrt{3t-11}\)
\(\Leftrightarrow\frac{\left[4-t\right]^2}{16}=3t-11\)
\(\Leftrightarrow t^2-56t+192=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=28+4\sqrt{37}\\t=28-4\sqrt{37}\end{cases}}\)
thế vào x+3=t suy ra
\(\orbr{\begin{cases}x=25+4\sqrt{37}\left[loại\right]\\x=25-4\sqrt{37}\left[nhận\right]\end{cases}}\)
\(S=\left\{25-4\sqrt{37}\right\}\)
