\(2x^3+x^2-8x-4=0\)

\(\Leftrightarrow\)\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)

\(\Leftrightarrow\)\(\left(2x+1\right)\left(x^2-4\right)=0\)

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

đến đây bạn làm tiếp nha

\(2x^3+x^2-8x-4=0\)

\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)

\(\left(x^2-4\right)\left(2x+1\right)=0\)

\(1.x^2-4=0\)

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

\(\Rightarrow x=\pm2\)

\(2.2x+1=0\)

\(x=-\frac{1}{2}\)

\(2x^3+x^2-8x-4=0\)

\(\Leftrightarrow x^2.\left(2x+1\right)-4\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\)2x + 1 = 0 => x = -1/2

Hoặc x - 2 = 0 => x = 2

Hoặc x + 2 = 0 => x = -2

Vậy x = -1/2 hoặc x = 2 hoặc x = -2

x=-2

X=2

\(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)

\(\Leftrightarrow8x+16=0\)

\(\Leftrightarrow x=-2\)

Ta co´:

             -4 ( 2x + 9 ) - ( -8x + 3 ) - ( x + 13 ) = 0

                                -8x - 36 + 8x -3 - x - 13 = 0

            .             ( -8x + 8x ) - 36 - 3 - x - 13 = 0

                                              - 36 - 3 - x - 13 = 0

                                                                     x = 0 + 36 + 3 + 13

                                                                     x = 52

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2-30x+11x-33=0\)

\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

Vậy \(x\in\left\{3;-\frac{11}{10}\right\}.\)

           Bài làm :

Ta có :

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2-30x+11x-33=0\)

\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

Vậy x=3 hoặc x=-11/10

Tử \(x^4+2x^3+8x+16\)

\(=x^4-2x^3+4x^2+4x^3-8x^2+16x+4x^2-8x+16\)

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

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

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

Mẫu \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

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

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

Thay tử và mẫu vào ta có:\(\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}\ge0\)

Dấu "=" khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Vậy Min=0 khi x=-2