a) Ta có: \(\sqrt{2x^2+x+9}+\sqrt{2x^2-x+1}=x+4\)
\(\Leftrightarrow2x^2+x+9+2x^2-x+1+2\cdot\sqrt{2x^2+x+9}\cdot\sqrt{2x^2-x+1}=x^2+8x+16\)
\(\Leftrightarrow4x^2+10+2\cdot\sqrt{4x^4+19x^2-8x+9}=x^2+8x+16\)
\(\Leftrightarrow4x^2+10+\sqrt{16x^4+76x^2-32x+36}-x^2-8x-16=0\)
\(\Leftrightarrow3x^2-8x-6+\sqrt{16x^4+76x^2-32x+36}=0\)
a/ Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+x+9}=a>0\\\sqrt{2x^2-x+1}=b>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=2\left(x+4\right)\)
Phương trình trở thành:
\(a+b=\frac{a^2-b^2}{2}\Leftrightarrow2\left(a+b\right)=\left(a+b\right)\left(a-b\right)\)
\(\Leftrightarrow a-b=2\Leftrightarrow a=b+2\)
\(\Leftrightarrow\sqrt{2x^2+x+9}=\sqrt{2x^2-x+1}+2\)
\(\Leftrightarrow2x^2+x+9=2x^2-x+1+4+4\sqrt{2x^2-x+1}\)
\(\Leftrightarrow x+2=2\sqrt{2x^2-x+1}\) (\(x\ge-2\))
\(\Leftrightarrow x^2+4x+4=4\left(2x^2-x+1\right)\)
\(\Leftrightarrow7x^2-8x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{8}{7}\end{matrix}\right.\)
b/
Do vế trái luôn dương nên vế phải dương \(\Rightarrow x>0\)
\(\Leftrightarrow2x-\sqrt{2x^2+x+1}+x-\sqrt{x^2-x+1}=0\)
\(\Leftrightarrow\frac{2x^2-x-1}{2x+\sqrt{2x^2+x+1}}+\frac{x-1}{x+\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+1\right)}{2x+\sqrt{2x^2+x+1}}+\frac{x-1}{x+\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2x+1}{2x+\sqrt{2x^2+x+1}}+\frac{1}{x+\sqrt{x^2-x+1}}\right)=0\)
\(\Leftrightarrow x-1=0\) (ngoặc phía sau luôn dương với \(x>0\))
\(\Rightarrow x=1\)
