Xét thấy x = 0 không thỏa mãn pt

Ta có : \(6x^4+7x^3-36x^2+7x+6=0\)

\(\Leftrightarrow x^2\left(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}\right)=0\)

\(\Leftrightarrow6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-36-12=0\)

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-48=0\)

Đặt \(x+\frac{1}{x}=a\)

\(pt\Leftrightarrow6a^2-7a-48=0\)

\(\Leftrightarrow6\left(a^2-\frac{7}{6}a-8\right)=0\)

\(\Leftrightarrow a^2-\frac{7}{6}a-8=0\)

\(\Leftrightarrow a^2-2\cdot a\cdot\frac{7}{12}+\frac{49}{144}-\frac{1201}{144}=0\)

\(\Leftrightarrow\left(a-\frac{7}{12}\right)^2=\left(\frac{\pm\sqrt{1201}}{12}\right)^2\)

\(\Leftrightarrow a=\frac{\pm\sqrt{1201}+7}{12}\)

\(\Leftrightarrow x+\frac{1}{x}=\frac{\pm\sqrt{1201}+7}{12}\)

Giải nốt nha bạn. Nghiệm hơi xấu

Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\)

\(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

Đặt \(x+\frac{1}{x}=a\) (\(\left|a\right|\ge2\)) \(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(6\left(a^2-2\right)+7a-36=0\)

\(\Leftrightarrow6a^2+7a-48=0\)

Nghiệm xấu

Ta có : \(6x^4+7x^3-36x^2-7x+6=0\)

\(6x^4-12x^3+19x^3-38x^2+2x^2-4x-3x+6=0\)

\(6x^3\left(x-2\right)+19x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)=0\)

\(\left(x-2\right)\left(6x^3+19x^2+2x-3\right)=0\)

\(\left(x-2\right)\left(6x^3+18x^2+x^2+3x-x-3\right)=0\)

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

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

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

Cách 2 : A = \(6x^4+7x^3-36x^2-7x+6=0\)

= \(6x^2\left(x^2+\dfrac{7}{6}x-6-\dfrac{7}{6x}+\dfrac{1}{x^2}\right)=0\)

= \(6x^2\left(\left(x^2+\dfrac{1}{x^2}-2\right)+\dfrac{7}{6}\left(x-\dfrac{1}{x}\right)-4\right)=0\)

= \(6x^2\left(\left(x-\dfrac{1}{x}\right)^2+\dfrac{7}{6}\left(x-\dfrac{1}{x}\right)-4\right)=0\)

Đặt \(A'=\left(x-\dfrac{1}{x}\right)^2+\dfrac{7}{6}\left(x-\dfrac{1}{x}\right)-4\)

và \(t=x-\dfrac{1}{x}\Rightarrow A'=t^2+\dfrac{7}{6}t-4\)

\(=t^2+2t.\dfrac{7}{12}+\left(\dfrac{7}{12}\right)^2-\left(\dfrac{7}{12}\right)^2-4\)

\(=\left(t+\dfrac{7}{12}\right)^2-\left(\dfrac{25}{12}\right)^2=\left(t-\dfrac{3}{2}\right)\left(t+\dfrac{8}{3}\right)\)

\(\Rightarrow A=6x^2\left(x-\dfrac{1}{x}-\dfrac{3}{2}\right)\left(x-\dfrac{1}{x}+\dfrac{8}{3}\right)=0\)

\(=6x^2\left(\dfrac{x^2-1-\dfrac{3}{2}x}{x}\right)\left(\dfrac{x^2-1+\dfrac{8}{3}x}{x}\right)=0\)

\(=\dfrac{6x^2\left(\left(x-\dfrac{3}{4}\right)^2-\left(\dfrac{5}{4}\right)^2\right)\left(\left(x+\dfrac{4}{3}\right)^2-\left(\dfrac{5}{3}\right)^2\right)}{x^2}=0\)

\(=6\left(x-2\right)\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{3}\right)\left(x+3\right)=0\)

Rồi giải ra các nghiệm trên nha !!! ( mình mỏi tay quá, phần sau các bạn làm như cách 1 nha )

.................. Làm sao nhỉ ??? Em ko thua đâu ! Đợi chút , em tìm cách giải !!!

(6x⁴-12x³)+(19³-38x²)+(2x²-4x)-(3x-6)=0

6x^3(x-2)+19x^2(x-2)+2x(x-2)-3(x-2)=0

(x-2)(6x^3+19x^2+2x-3)=0

(x-2)[(6x^3+18x^2)+(x^2+3x)-(x+3)]=0

(x-2)(x+3)(6x^2+x-1)=0

(x-2)(x+3)[(6x^2+3x)-(2x+1)]=0

(x-2)(x+3)(2x+1)(3x-1)=0

⇒ x=2

x=-3

x=-1/2

x=1/3