ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{4x+5}=a\ge0\\\sqrt{x^2+2x+5}=b>0\end{matrix}\right.\)
\(\Rightarrow\left(a^2-3\right)b=\left(b^2-3\right)a\)
\(\Leftrightarrow a^2b-ab^2+3\left(a-b\right)=0\)
\(\Leftrightarrow ab\left(a-b\right)+3\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(ab+3\right)=0\)
\(\Leftrightarrow a=b\Rightarrow x^2+2x+5=4x+5\)
\(\Leftrightarrow x^2-2x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Đặt \(a=\sqrt{x^2+2x+5};b=\sqrt{4x+5}\left(ab\ge0\right)\) , ta có :
( b2 - 3 ) a = ( a2 - 3 ) b
⇔ ab2 - 3a = a2b - 3b
⇔ ab ( b - a ) + 3 ( b - a ) = 0
⇔ ( b - a ) ( ab + 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}b-a=0\\ab+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=a\\ab=-3\end{matrix}\right.\)
Mà ab ≥ 0 nên :
\(a=b\Rightarrow\sqrt{x^2+2x+5}=\sqrt{4x+5}\)
\(\Leftrightarrow x^2+2x+5=4x+5\)
\(\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(ĐK:x\ge\frac{-5}{4}\)
Dễ thấy \(x=\frac{-1}{2};x=\frac{-5}{4}\)không là nghiệm của phương trình nên:
\(\left(4x+2\right)\sqrt{x^2+2x+5}=\left(x^2+2x+2\right)\sqrt{4x+5}\Leftrightarrow\sqrt{\frac{x^2+2x+5}{4x+5}}=\frac{x^2+2x+2}{4x+2}\)\(\Leftrightarrow\sqrt{\frac{x^2+2x+5}{4x+5}}-1=\frac{x^2+2x+2}{4x+2}-1\Leftrightarrow\frac{\frac{x^2+2x+5}{4x+5}-1}{\sqrt{\frac{x^2+2x+5}{4x+5}}+1}=\frac{x^2-2x}{4x+2}\)\(\Leftrightarrow\frac{x^2-2x}{\left(4x+5\right)\left(\sqrt{\frac{x^2+2x+5}{4x+5}}+1\right)}=\frac{x^2-2x}{4x+2}\)\(\Leftrightarrow\left(x^2-2x\right)\left(\frac{1}{\left(4x+5\right)\left(\sqrt{\frac{x^2+2x+5}{4x+5}}+1\right)}-\frac{1}{4x+2}\right)=0\)
* Th1: \(x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow x\in\left\{0;2\right\}\)
* Th2: \(\frac{1}{\left(4x+5\right)\left(\sqrt{\frac{x^2+2x+5}{4x+5}}+1\right)}=\frac{1}{4x+2}\)hay \(\left(4x+5\right)\left(\sqrt{\frac{x^2+2x+5}{4x+5}}+1\right)=4x+2\)\(\Leftrightarrow\sqrt{\frac{x^2+2x+5}{4x+5}}+1=\frac{4x+2}{4x+5}\Leftrightarrow\sqrt{\frac{x^2+2x+5}{4x+5}}=\frac{-3}{4x+5}\left(\Phi\right)\)
Xét \(VT=\sqrt{\frac{x^2+2x+5}{4x+5}}>0\forall x>\frac{-5}{4}\)mà \(VP=\frac{-3}{4x+5}< 0\forall x>\frac{-5}{4}\)nên \(\left(\Phi\right)\)vô nghiệm
Vậy tập nghiệm của phương trình là \(S=\left\{0;2\right\}\)
