ĐK: \(\forall x\in R\)
\(PT\Leftrightarrow\left(7x^2-10x+14\right)^2=25\left(x^4+4\right)\)
\(\Leftrightarrow49x^4-140x^3+296x^2-280x+196=25x^4+100\)
\(\Leftrightarrow24x^4-140x^3+296x^2-280x+96=0\)
\(\Leftrightarrow6x^4-35x^3+74x^2-70x+24=0\)
Với \(x=0\) => Không thỏa mãn phương trình.
Với \(x\ne0\) , chia cả 2 vế cho \(x^2\):
\(\Leftrightarrow6x^2-35x+74-\frac{70}{x}+24x^2=0\)
\(\Leftrightarrow6\left(x^2+\frac{4}{x^2}\right)-35\left(x+\frac{2}{x}\right)+74=0\)
Đặt \(x+\frac{2}{x}=t\) \(\Rightarrow t^2=x^2+\frac{4}{x^2}+4\)
\(\Leftrightarrow6\left(t^2-4\right)-35t+74=0\) \(\Leftrightarrow6t^2-35t+50=0\Rightarrow\left[{}\begin{matrix}t=\frac{10}{3}\\t=\frac{5}{2}\end{matrix}\right.\)
+\(t=\frac{10}{3}\Rightarrow x+\frac{2}{x}=\frac{10}{3}\Rightarrow3x^2-10x+6=0\) \(\Rightarrow x=\frac{5\pm\sqrt{7}}{3}\)
+\(t=\frac{5}{2}\Rightarrow x+\frac{2}{x}=\frac{5}{2}\Rightarrow2x^2-5x+4=0\) (Vô no)
Vậy...
