2) c) Đặt \(\left(\sqrt{x+1};\sqrt{x+6}\right)=\left(a;b\right)\ge0\)
Ta có hệ phương trình
\(\left\{{}\begin{matrix}a+b=5\\b^2-a^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+b=5\\\left(b-a\right)\left(b+a\right)=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+b=5\\\left(b-a\right).5=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+b=5\\b-a=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+6}=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+1=4\left(x\ge-1\right)\\x+6=9\left(x\ge-6\right)\end{matrix}\right.\)
\(\Rightarrow x=3\)
Vậy: Phương trình có nghiệm duy nhất x=3
\(\sqrt{x+1}+\sqrt{x+6}=5\)
đkxđ : \(\)\(\left\{{}\begin{matrix}x>-1\\x>-6\end{matrix}\right.\Rightarrow x>-1}\)
bình phương 2 vế ta có :
\(x+1+2\sqrt{\left(x+1\right)\left(x+6\right)}+x+6=5\)
\(\Leftrightarrow2x+2=-2\sqrt{x^2+7x+6}\)
đk:\(2x+2>0\Leftrightarrow x>-1\)
bình phương 2 vế ta có :
\(4x^2+8x+4=4x^2+28x+24\)
\(\Leftrightarrow20x+20=0\Leftrightarrow x=-1\left(KTM\right)\)
vậy pt vô nghiệm
