1, \(ĐKXĐ:x^3+2x^2+2x+1\ge0\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\ge0\Leftrightarrow x\ge-1\)
\(9x^2+2x+2+2\sqrt{x^3+2x^2+2x+1}=0\)
\(\Leftrightarrow9x^2+2x+2+2\sqrt{\left(x+1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow8x^2+x+1+2\sqrt{\left(x+1\right)\left(x^2+x+1\right)}+x^2+x+1=0\)
\(\Leftrightarrow8x^2+\left(\sqrt{x+1}+\sqrt{x^2+x+1}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x^2=0\\\sqrt{x+1}+\sqrt{x^2+x+1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x+1}+\sqrt{x^2+x+1}=0\end{matrix}\right.\)
- Thay \(x=0\) vào \(\sqrt{x+1}+\sqrt{x^2+x+1}\) ta có:
\(\sqrt{x+1}+\sqrt{x^2+x+1}=\sqrt{0+1}+\sqrt{0+0+1}=2\ne0\)
\(\Rightarrow8x^2,\sqrt{x+1}+\sqrt{x^2+x+1}\) không thể cùng đồng thời bằng 0.
\(\Rightarrow\)Pt vô nghiệm.
\(2,,\sqrt{2x^2+12x+5}+\sqrt{2x^2-3x+5}=8\sqrt{x}\)\(\left(1\right)\)
\(đặt:\sqrt{2x^2+12x+5}=a\ge0;\sqrt{2x^2-3x+5}=b\ge0\Rightarrow a^2-b^2=15x\Rightarrow\sqrt{x}=\sqrt{\dfrac{a^2-b^2}{15}}\)
\(\Rightarrow pt\left(1\right)\Leftrightarrow a+b=8\sqrt{\dfrac{a^2-b^2}{15}}\Leftrightarrow a+b-\dfrac{8}{\sqrt{15}}\sqrt{\left(a+b\right)\left(a-b\right)}=0\Leftrightarrow\sqrt{a+b}\left[\sqrt{a+b}-\dfrac{8}{\sqrt{15}}\sqrt{a-b}\right]=0\Leftrightarrow\left[{}\begin{matrix}a=-b\left(loại\right)do:\sqrt{2x^2+12x+5}=-\sqrt{2x^2-3x+5}< 0\\\sqrt{a+b}-8\sqrt{\dfrac{a-b}{15}}=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\sqrt{a+b}=8\sqrt{\dfrac{a-b}{15}}\Leftrightarrow a+b=\dfrac{64}{15}\left(a-b\right)\Rightarrow a=\dfrac{79}{49}b\Rightarrow\sqrt{2x^2+12x+5}=\dfrac{79}{49}\sqrt{2x^2-3x+5}\Rightarrow x=.....\)
\(1.\) \(xem\) \(lại\) \(do\) \(pt\Leftrightarrow8x^2=-\left(\sqrt{x+1}+\sqrt{x^2+x+1}\right)^2< 0\)
