đk x\(\ge-1\)
pt \(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=25\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=25\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=21-3x\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(21-3x\right)^2\)đk \(x\le7\)
\(\Leftrightarrow8x^2+20x+12=9x^2-126x+441\)
\(\Leftrightarrow x^2-146x+429=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=143\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)
ĐK : \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge-1\)
Ta có :
\(\sqrt{x+1}+\sqrt{2x+3}=5\\ \Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=25\\ \Leftrightarrow x+1+2x+3+2\sqrt{\left(x+1\right)\left(2x+3\right)}=25\\ \Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=25-3x-4\\ \Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=21-3x\\ \Leftrightarrow4\left(x+1\right)\left(2x+3\right)=\left(21-3x\right)^2\\ \Leftrightarrow4\left(2x^2+5x+3\right)=441-126x+9x^2\\ \Leftrightarrow8x^2+20x+12=441-126x+9x^2\\ \Leftrightarrow441-126x+9x^2-8x^2-20x-12=0\\ \Leftrightarrow x^2-146x+429=0\\ \Leftrightarrow\left[{}\begin{matrix}x=143\left(TMĐK\right)\\x=3\left(TMĐK\right)\end{matrix}\right.\)
Vậy phương trình đã cho có 2 nghiệm là x=143 và x=3
