Ta có: \(\sqrt{x^2+2x+3}+\sqrt{x^2+x+2}=2x+2\)
Bình phương 2 vế ta có:
\(2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=4\left(x+1\right)^2-x^2-2x-3-x^2-x-2\) (\(x\ge-1\))
\(\Leftrightarrow2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=4x^2+8x+4-2x^2-3x-5\)
\(\Leftrightarrow2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=2x^2+5x-1\)\(\Leftrightarrow2\sqrt{\left(x^2+2x+3\right)\left(x^2+x+2\right)}=2x^2+5x-1\)
Bình phương 2 vế, ta được:
\(4\left(x^2+2x+3\right)\left(x^2+x+2\right)=\left(2x^2+5x-1\right)^2\) ( ĐK:\(\left[{}\begin{matrix}x\le\dfrac{-5-\sqrt{33}}{4}\\x\ge\dfrac{-5+\sqrt{33}}{4}\end{matrix}\right.\))
\(\Leftrightarrow4\left(x^4+x^3+2x^2+2x^3+2x^2+4x+3x^2+3x+6\right)=4x^4+20x^3+21x^2-10x+1\)
\(\Leftrightarrow4x^4+4x^3+8x^2+8x^3+8x^2+16x+12x^2+12x+24=4x^4+20x^3+21x^2-10x+1\)\(\Leftrightarrow-8x^3+7x^2+38x+23=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{8}\\x=-1\left(loai\right)\end{matrix}\right.\)
Vậy nghiệm của PT là \(x=\dfrac{23}{8}\)
Đặt \(\sqrt{x^2+2x+3}=a;\sqrt{x^2+x+2}=b\) ĐK : \(a;b>0\)
PT <=> a + b = 2(a2 - b2)
<=> a + b = 2(a - b)(a + b)
<=> (a + b)(2a - 2b - 1) = 0
<=> \(\left[{}\begin{matrix}a+b=0\\2a=2b+1\end{matrix}\right.\Leftrightarrow2a=2b+1\left(\text{vì a ; b > 0}\right)\)
Khi đó \(2\sqrt{x^2+2x+3}=2\sqrt{x^2+x+2}+1\)
\(\Leftrightarrow4\left(x^2+2x+3\right)=4\left(x^2+x+2\right)+4\sqrt{x^2+x+2}+1\)
<=> \(4\sqrt{x^2+x+2}=4x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}16\left(x^2+x+2\right)=16x^2+24x+9\\x\ge-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x=23\\x\ge-\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x=\dfrac{23}{8}\)
