ĐKXĐ : \(\left\{{}\begin{matrix}x+9\ge0\\2x+4\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-9\\2x\ge-4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-9\\x\ge-2\end{matrix}\right.\)
<=> \(x\ge-2\)
Ta có : \(\sqrt{x+9}=5-\sqrt{2x+4}\)
<=> \(\sqrt{x+9}+\sqrt{2x+4}=5\)
<=> \(\left(\sqrt{x+9}+\sqrt{2x+4}\right)^2=5^2\)
<=> \(\left(x+9\right)+2\sqrt{\left(x+9\right)\left(2x+4\right)}+\left(2x+4\right)=25\)
ĐKXĐ : \(x\le4\)
=> \(-2\le x\le4\)
<=> \(x+9+2\sqrt{\left(x+9\right)\left(2x+4\right)}+2x+4=25\)
<=> \(2\sqrt{\left(x+9\right)\left(2x+4\right)}=25-x-9-2x-4\)
<=> \(2\sqrt{\left(x+9\right)\left(2x+4\right)}=12-3x\)
<=> \(\left(2\sqrt{\left(x+9\right)\left(2x+4\right)}\right)^2=\left(12-3x\right)^2\)
<=> \(4\left(x+9\right)\left(2x+4\right)=\left(12-3x\right)^2\)
<=> \(4\left(2x^2+18x+4x+36\right)=144-72x+9x^2\)
<=> \(8x^2+72x+16x+144=144-72x+9x^2\)
<=> \(8x^2+72x+16x+144-9x^2-144+72x=0\)
<=> \(-x^2+160x=0\)
<=> \(x\left(160-x\right)=0\)
<=> \(\left\{{}\begin{matrix}x=0\\160-x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\left(TM\right)\\x=160\left(L\right)\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là x = 0 .