Question

giải pt :

\(^{^{ }6x^4}\)- \(^{35x^3}\)+ \(^{62x^2}\) - \(35x\) +6 = 0

Answer 1

Ta có : \(6x^4-35x^3+62x^2-35x+6=0\)

=> \(6x^4-3x^3-32x^3+16x^2+46x^2-23x-12x+6=0\)

=> \(3x^3\left(2x-1\right)-16x^2\left(2x-1\right)+23x\left(2x-1\right)-6\left(2x-1\right)=0\)

=> \(\left(3x^3-16x^2+23x-6\right)\left(2x-1\right)=0\)

=> \(\left(3x^3-x^2-15x^2+5x+18x-6\right)\left(2x-1\right)=0\)

=> \(\left(x^2\left(3x-1\right)-5x\left(3x-1\right)+6\left(3x-1\right)\right)\left(2x-1\right)=0\)

=> \(\left(x^2-5x+6\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left(x^2-2x-3x+6\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left(x\left(x-2\right)-3\left(x-2\right)\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left(x-3\right)\left(x-2\right)\left(3x-1\right)\left(2x-1\right)=0\)

=> \(\left[{}\begin{matrix}x-3=0\\x-2=0\\3x-1=0\\2x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\x=2\\x=\frac{1}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{2,3,\frac{1}{2},\frac{1}{3}\right\}\)

Answer 2

Nhận thấy \(x=0\) ko là nghiệm, chia 2 vế của pt cho \(x^2\)

\(6x^2+\frac{6}{x^2}-35x-\frac{35}{x}+62=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)+62=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(6\left(t^2-2\right)-35t+62=0\)

\(\Leftrightarrow6t^2-35t+50=0\Rightarrow\left[{}\begin{matrix}t=\frac{5}{2}\\t=\frac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\x+\frac{1}{x}=\frac{10}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-5x+2=0\\3x^2-10x+3=0\end{matrix}\right.\)