ĐKXĐ : \(\left\{{}\begin{matrix}3x-2\ge0\\3+x\ge0\\5x+4\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge\frac{2}{3}\\x\ge-3\\x\ge-\frac{4}{5}\end{matrix}\right.\)
=> \(x\ge\frac{2}{3}\) (1)
Ta có : \(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)
<=> \(\left(\sqrt{3x-2}+\sqrt{3+x}\right)^2=\left(\sqrt{5x+4}\right)^2\)
<=> \(\left(3x-2\right)+2\sqrt{\left(3x-2\right)\left(3+x\right)}+\left(3+x\right)=5x+4\)
<=> \(3x-2+2\sqrt{\left(3x-2\right)\left(3+x\right)}+3+x=5x+4\)
<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=5x+4+2-3-x-3x\)
<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=x+3\)
<=> \(\sqrt{\left(3x-2\right)\left(3+x\right)}=\frac{x+3}{2}\)
ĐKXĐ : \(\frac{x+3}{2}\ge0\)
=> \(x+3\ge0\)
=> \(x\ge-3\) (2)
Từ (1) và (2)
=> \(x\ge\frac{2}{3}\)
<=> \(\left(\sqrt{\left(3x-2\right)\left(3+x\right)}\right)^2=\left(\frac{x+3}{2}\right)^2\)
<=> \(\left(3x-2\right)\left(3+x\right)=\frac{\left(x+3\right)^2}{4}\)
<=> \(9x-6+3x^2-2x=\frac{x^2+6x+9}{4}\)
<=> \(\frac{4\left(9x-6+3x^2-2x\right)}{4}=\frac{x^2+6x+9}{4}\)
<=> \(4\left(9x-6+3x^2-2x\right)=x^2+6x+9\)
<=> \(36x-24+12x^2-8x=x^2+6x+9\)
<=> \(36x-24+12x^2-8x-x^2-6x-9=0\)
<=> \(22x-33+11x^2=0\)
<=> \(11x^2+33x-11x-33=0\)
<=> \(11x\left(x-1\right)+33\left(x-1\right)=0\)
<=> \(\left(11x+33\right)\left(x-1\right)=0\)
<=> \(\left\{{}\begin{matrix}11x+33=0\\x-1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=-3\left(L\right)\\x=1\left(TM\right)\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là x = 1 .