a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)

\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\)

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

hay \(x\in\left\{-2;1\right\}\)

b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)

hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)

Ta có: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3-3x^2+16x^2-8x-6x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[3x^2\left(2x-1\right)+8x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+8x-3\right)=0\)

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

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

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

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

Vậy: \(S=\left\{-2;\dfrac{1}{2};-3;\dfrac{1}{3}\right\}\)

1)\(25x+3\left(4-6x\right)=50\)

\(25x+12-18x=50\)

\(7x+12=50\)

\(7x=38\)

\(x=\frac{38}{7}\)

2)\(4\left(2x+3\right)+2\left(3x+1\right)=120\)

\(8x+12+6x+2=120\)

\(14x+14=120\)

\(14x=106\)

\(x=\frac{53}{7}\)

3)\(2\left(3x+2\right)+3\left(4x+1\right)=200\)

\(6x+4+12x+3=200\)

\(18x+7=200\)

\(18x=193\)

\(x=\frac{193}{18}\)

Bạn dùng phương pháp phân tích đa thức thành nhân tử sẽ đc :

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

<=> x=1/2 hoặc x=-3 hoặc x=-2 hoặc x=1/3

Vậy .............

Tk mk nha

giải ra

             \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow\)\(6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

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

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

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

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

P/S:lm tiếp nha   x = -2;  x = -3;  x = 1/2;  x = 1/3

\(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow\)  \(6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\)  \(6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

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

\(\Leftrightarrow\)  \(\left(x+2\right)\left[6x^2\left(x+3\right)-5x\left(x+3\right)+x+3\right]=0\)

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

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

\(\Leftrightarrow\)   \(x+2=0\)  hoặc  \(x+3=0\)  hoặc  \(2x-1=0\)  hoặc  \(3x-1=0\)

\(\Leftrightarrow\)   \(x=-2\)  hoặc \(x=-3\)  hoặc  \(x=\frac{1}{2}\)  hoặc  \(x=\frac{1}{3}\)

Vậy, tập nghiệm của pt là  \(S=\left\{-2;-3;\frac{1}{2};\frac{1}{3}\right\}\)

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

\(6\left(x^2+\frac{1}{x^2}\right)+25\left(x-\frac{1}{x}\right)+12=0\)

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

\(\Rightarrow6\left(t^2+2\right)+25t+12=0\)

\(\Leftrightarrow6t^2+25t+24=0\Rightarrow\left[{}\begin{matrix}t=-\frac{3}{2}\\t=-\frac{8}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{x}=-\frac{3}{2}\\x-\frac{1}{x}=-\frac{8}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-2=0\\3x^2+8x-3=0\end{matrix}\right.\)

72:6=(48+24):6=48:6 + 24: 6 = 8+4=12