ĐKXĐ : \(x-1\ge0\)
=> \(x\ge1\)
\(3x-5\ge0\)
=> \(x\ge-0,6\)
Vậy ĐKXĐ là \(x\ge1\)
Ta có : \(2+\sqrt{x-1}+\sqrt{3x-5}=x\)
<=> \(\sqrt{x-1}+\sqrt{3x-5}=x-2\)
<=> \(\left(\sqrt{x-1}+\sqrt{3x-5}\right)^2=\left(x-2\right)^2\)
<=> \(\left(|x-1|\right)+2\sqrt{\left(x-1\right)\left(3x-5\right)}+\left(|3x-5|\right)=x^2-4x+4\)<=> \(\left(x-1\right)+2\sqrt{3x^2-3x-5x+5}+\left(3x-5\right)=x^2-4x+4\)<=> \(x-1+2\sqrt{3x^2-8x+5}+3x-5-x^2+4x-4=0\)<=> \(8x-10-x^2+2\sqrt{3x^2-8x+5}=0\)<=>\(8x-10-x^2+2\sqrt{3x^2-8x+5}=0\)<=>
\(2\sqrt{3x^2-8x+5}=10-8x+x^2\)
<=>\(\left(2\sqrt{3x^2-8x+5}\right)^2=\left(10+x^2-8x\right)^2\)<=> \(4\left(3x^2-8x+5\right)=100+x^4+64x^2-160x-16x^3+20x^2\)
<=> \(12x^2-32x+20-100-x^4-64x^2+160x+16x^3-20x^2=0\)
<=> \(-x^4+16x^3-72x^2+128x-80=0\)
<=> \(-x^4+2x^3+14x^3-28x^2-44x^2+88x+40x-80=0\)<=> \(-x^3\left(x-2\right)+14x^2\left(x-2\right)-44x\left(x-2\right)+40\left(x-2\right)=0\)
<=> \(-\left(x-2\right)\left(x^3-14x^2+44x-40\right)=0\)
<=> \(-\left(x-2\right)\left(x^3-2x^2-12x^2+24x+20x-40\right)=0\)
<=> \(-\left(x-2\right)\left(x^2\left(x-2\right)-12x\left(x-2\right)+20\left(x-2\right)\right)=0\)
<=> \(-\left(x-2\right)\left(x-2\right)\left(x^2-12x+20\right)=0\)
<=> \(-\left(x-2\right)\left(x-2\right)\left(x^2-2x-10x+20\right)=0\)
<=> \(-\left(x-2\right)\left(x-2\right)\left(x-2\right)\left(x-10\right)=0\)
<=> \(-\left(x-2\right)\left(x-2\right)^2\left(x-10\right)=0\)
<=> \(-\left(x-2\right)^3\left(x-10\right)=0\)
<=> \(\left(x-2\right)^3\left(x-10\right)=0\)
<=> \(\left\{{}\begin{matrix}\left(x-2\right)^3=0\\x-10=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=2\\x=10\end{matrix}\right.\)( TM )
Vậy phương trình trên có nghiệm là x = 2, 10 .
