DKXD: x\(\ge1\)
Ta có: \(2x^2+5x-1=7\sqrt{x^3-1}\)\(\Leftrightarrow\left(2x^2+2x+2\right)+\left(3x-3\right)=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)
Do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{1}{4}>0\forall x\)
Nen ta chia hai ve cua phuong trinh cho \(x^2+x+1,\)ta duoc
\(2+3\times\frac{x-1}{x^2+x+1}=7\sqrt{\frac{x-1}{x^2+x+1}}\)
Dat \(\sqrt{\frac{x-1}{x^2+x+1}}=t\)\(\left(t\ge0\right)\)ta có
\(2+3t^2=7t\Leftrightarrow3t^2-7t+2=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=2\\t=\frac{1}{3}\end{cases}}\)
+) \(t=2\Rightarrow\frac{x-1}{x^2+x+1}=4\Rightarrow4x^2+3x+5=0\)
\(\left(ptvn\right)\)
+) \(t=\frac{1}{3}\Rightarrow\frac{x-1}{x^2+x+1}=\frac{1}{9}\)
TT bạn tu tinh nhé
