\(\dfrac{3x+3}{\sqrt{x}}=4+\dfrac{x+1}{\sqrt{x^2-x+1}}\) ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{3x+3}{\sqrt{x}}-6-\left(\dfrac{x+1}{\sqrt{x^2-x+1}}-2\right)=0\)
\(\Leftrightarrow\dfrac{3x+3-6\sqrt{x}}{\sqrt{x}}-\dfrac{x+1-2\sqrt{x^2-x+1}}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(x-2\sqrt{x}+1\right)}{\sqrt{x}}-\dfrac{x^2-x+1-2\sqrt{x^2-x+1}+1-x^2+2x-1}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-1\right)^2}{\sqrt{x}}-\dfrac{\left(\sqrt{x^2-x+1}-1\right)^2-\left(x-1\right)^2}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-1\right)^2}{\sqrt{x}}-\dfrac{\left(\dfrac{x\left(x-1\right)}{\sqrt{x^2-x+1}+1}\right)^2-\left(x-1\right)^2}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-1\right)^2}{\sqrt{x}}-\left(x-1\right)^2\dfrac{\left(\dfrac{x}{\sqrt{x^2-x+1}+1}\right)^2-1}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\left[\dfrac{3}{\sqrt{x}}-\left(\sqrt{x}+1\right)^2\dfrac{\left(\dfrac{x}{\sqrt{x^2-x+1}+1}\right)^2-1}{\sqrt{x^2-x+1}}\right]=0\)
Ta có \(\dfrac{3}{\sqrt{x}}-\left(\sqrt{x}+1\right)^2\dfrac{\left(\dfrac{x}{\sqrt{x^2-x+1}+1}\right)^2-1}{\sqrt{x^2-x+1}}\) vô nghiệm
Vậy x=1
