x^3 - 4x - 3x - 6 =0
\(\Leftrightarrow\)x(x^2-4) - 3(x+2) =0
\(\Leftrightarrow\)x(x-2)(x+2) - 3(x+2)=0
\(\Leftrightarrow\)(x+2)(x^2-2x-3)=0
\(\Leftrightarrow\)(x+2)(x^2-3x+x-3)=0
\(\Leftrightarrow\)(x+2)(x+1)(x-3)=0
\(\Leftrightarrow\)x=3 hoặc x=-1 hoặc x=-3
Gọi \(A\) = \(x^3-7x-6\)
\(A=x^3-x^2+x^2-x-6x+6\)
\(A=x^2\left(x-1\right)+x\left(x-1\right)-6\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2+x-6\right)\)
\(A=\left(x-1\right)\left(x^2+3x-2x-6\right)\)
\(A=\left(x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(A=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
x3 - 7x - 6 = 0
=> x3 + 2x2 + x2 + 2x - 3x2 - 6x - 3x - 6 = 0
=> x2 (x + 2) + x .(x + 2) - 3x(x + 2) - 3(x + 2) = 0
=> (x + 2) .(x2 + x - 3x - 3) = 0
=> (x + 2) . [x (x + 1) - 3(x + 1)] = 0
=> (x + 2)(x + 1)(x - 3) = 0
=> x + 2 = 0 => x = -2
hoặc x + 1 = 0 => x = -1
hoặc x - 3 = 0 => x = 3
Vậy x = 3, x = -1, x = -2
Ta có :
\(x^3-7x-6=0\)
\(=x^3+x^2-x^2-6x-x-6=0\)
\(=\left(x^3+x^2\right)-\left(x^2+x\right)-\left(6x+6\right)=0\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)=0\)
\(=\left(x^2-x-6\right)\left(x+1\right)=0\)
\(=\left(x^2-3x+2x-6\right)\left(x+1\right)=0\)
\(=\left[x\left(x-3\right)+2\left(x-3\right)\right]\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow x+2=0\)hoặc \(x-3=0\)hoặc \(x+1=0\)
\(\Leftrightarrow x\in\left\{-2;-1;3\right\}\)
