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

\(PT\Leftrightarrow\left(x^4-x^3\right)-\left(6x^3-6x^2\right)+\left(12x^2-12x\right)-\left(9x-9\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)-6x^2\left(x-1\right)+12x\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-3x^2\right)-\left(3x^2-9x\right)+\left(3x-9\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-3x\left(x-3\right)+3\left(x-3\right)\right]=0\)

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

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

Vậy..

a: \(\left(3x-2\right)\cdot\left(\dfrac{2}{7}\left(x+3\right)-\dfrac{4x-3}{5}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\dfrac{2}{7}x+\dfrac{6}{7}-\dfrac{4}{5}x+\dfrac{3}{5}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(-\dfrac{18}{35}x+\dfrac{51}{35}\right)=0\)

=>x=2/3 hoặc x=51/18=17/6

b \(\left(3.3-11x\right)\left(\dfrac{7x+2}{5}+\dfrac{2\left(1-3x\right)}{3}\right)=0\)

\(\Leftrightarrow\left(-10x+3\right)\left(21x+6+10-30x\right)=0\)

\(\Leftrightarrow\left(-10x+3\right)\left(-9x+16\right)=0\)

=>x=3/10 hoặc x=16/9

c: \(\dfrac{3}{7x-1}=\dfrac{1}{7x\left(3x-7\right)}\)

=>21x(3x-7)=7x-1

\(\Leftrightarrow63x^2-154x+1=0\)

\(\text{Δ}=\left(-154\right)^2-4\cdot63=23464\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{154-\sqrt{23464}}{126}\\x_2=\dfrac{154+\sqrt{23464}}{126}\end{matrix}\right.\)

a) Ta có: \(f\left(x\right)=x\left(x^2+x-2\right)=x\left(x-1\right)\left(x+2\right)\)

  Lập bảng xét dấu 

Vậy để \(f\left(x\right)>0\) \(\Leftrightarrow x\in\left(-2;0\right)\cup\left(1;+\infty\right)\)

b) Ta có: \(\left(3x^2+7x-6\right)\left(5x+8\right)^2\le0\)

\(\Leftrightarrow3x^2+7x-6\le0\) \(\Leftrightarrow-3\le x\le\dfrac{2}{3}\)

  Vậy \(x\in\left[-3;\dfrac{2}{3}\right]\)