ĐK: \(x\ge1\)
\(\sqrt{5x-1}-\sqrt{3x-2}=\sqrt{x-1}\)
\(\Leftrightarrow5x-1-2\sqrt{\left(5x-1\right)\left(3x-2\right)}+3x-2=x-1\)
\(\Leftrightarrow7x-4-2\sqrt{\left(5x-1\right)\left(3x-2\right)}=0\)
\(\Leftrightarrow7x-4=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow49x^2-56x+16=4\left(15x^2-13x+2\right)\)
\(\Leftrightarrow-11x^2-4x+8=0\)
\(\Leftrightarrow-11\left(x^2+\frac{4}{11}-\frac{8}{11}\right)=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{2}{11}+\frac{4}{121}-\frac{92}{121}=0\)
\(\Leftrightarrow\left(x+\frac{2}{11}\right)^2=\frac{92}{121}=\left(\frac{\pm\sqrt{92}}{11}\right)^2\)
\(\Leftrightarrow x=\frac{\pm\sqrt{92}-2}{11}\)( không thỏa ĐK )
Vậy pt vô nghiệm
Trần Phương Linh làm lại nhé, xin lỗi :D
Sửa dấu tương đương thứ 2:
\(\Leftrightarrow7x-2-2\sqrt{\left(5x-1\right)\left(3x-1\right)}=0\)
\(\Leftrightarrow7x-2=2\sqrt{\left(5x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow49x^2-28x+4=4\left(15x^2-13x+2\right)\)
\(\Leftrightarrow-11x^2+24x-4=0\)
\(\Leftrightarrow-11x^2+22x+2x-4=0\)
\(\Leftrightarrow-11x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-11x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(thoa\right)\\x=\frac{2}{11}\left(ko-thoa\right)\end{matrix}\right.\)
Vậy...
Điều kiện: \(\left\{ \begin{array}{l} 5x - 1 \ge 0\\ 3x - 2 \ge 0\\ x - 1 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge \dfrac{1}{5}\\ x \ge \dfrac{2}{3}\\ x \ge 1 \end{array} \right. \Leftrightarrow x \ge 1\)
Phương trình tương đương:
\(\sqrt{5x-1}=\sqrt{x-1}+\sqrt{3x-2}\\ \Leftrightarrow5x-1=x-1+3x-2+2\sqrt{\left(x-1\right)\left(3x-2\right)}\\ \Leftrightarrow x+2=2\sqrt{\left(x-1\right)\left(3x-2\right)}\)
Điều kiện: \(x+2\ge0\Leftrightarrow x\ge-2\)
\(\Leftrightarrow\left(x+2\right)^2=4\left(x-1\right)\left(3x-2\right)\\ \Leftrightarrow x^2+4x+4=4\left(3x^2-5x+2\right)\\ \Leftrightarrow x^2+4x+4=12x^2-20x+8\\ \Leftrightarrow11x^2-24x+4=0\\ \Leftrightarrow\left(x-2\right)\left(11x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\11x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=\frac{2}{11}\left(KTM\right)\end{matrix}\right.\)
Vậy \(x=2\)
