ĐKXĐ: \(x\ge\frac{1}{2}\)
\(7\left(\sqrt{3x+1}-\sqrt{2x-1}\right)=x+2\)
\(\Leftrightarrow7.\frac{3x+1-2x+1}{\sqrt{3x+1}+\sqrt{2x-1}}=x+2\)
\(\Leftrightarrow7.\frac{x+2}{\sqrt{3x+1}+\sqrt{2x-1}}-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{7}{\sqrt{3x+1}+\sqrt{2x-1}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\frac{7}{\sqrt{3x+1}+\sqrt{2x-1}}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(l\right)\\\sqrt{3x+1}+\sqrt{2x-1}=7\end{matrix}\right.\)
\(\Leftrightarrow5x+2\sqrt{\left(3x+1\right)\left(2x-1\right)}=49\)
\(\Leftrightarrow2\sqrt{6x^2-x-1}=49-5x\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(6x^2-x-1\right)=\left(49-5x\right)^2\\49-5x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-486x+2405=0\\x\le\frac{49}{5}\end{matrix}\right.\Leftrightarrow x=5\left(tm\right)\)
Vậy phương trình có nghiệm \(x=5\)
Cách khác:
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x+1}=a\\\sqrt{2x-1}=b\end{matrix}\right.\left(a>0;b\ge0\right)\)
\(pt\Leftrightarrow7\left(a-b\right)=a^2-b^2\)
\(\Leftrightarrow7\left(a-b\right)=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=7\end{matrix}\right.\)
TH1: \(a=b\)
\(\Leftrightarrow\sqrt{3x+1}=\sqrt{2x-1}\)
\(\Leftrightarrow3x+1=2x-1\)
\(\Leftrightarrow x=-2\left(l\right)\)
TH2: \(a+b=7\)
\(\Leftrightarrow\sqrt{3x+1}+\sqrt{2x-1}=7\)
\(\Leftrightarrow5x+2\sqrt{\left(3x+1\right)\left(2x-1\right)}=7\)
Giải tiếp giống cách kia.
