Ta có: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)
<=> \(\left(x^2+x\right)\left(x+6\right)-x^3=5x\)
<=> \(x^3+7x^2+6x-x^3=5x\)
<=> \(7x^2+x=0\)
<=> \(x\left(7x+1\right)=0\)
<=> \(\left[\begin{array}{nghiempt}x=0\\7x+1=0\end{array}\right.\)<=>\(\left[\begin{array}{nghiempt}x=0\\x=-\frac{1}{7}\end{array}\right.\)
Vậy x\(\in\left\{-\frac{1}{7};0\right\}\)
\(x\left(x+1\right)\left(x+6\right)=\left(x^2+x\right)\left(x+6\right)=x^3+6x^2+x^2+6x=x^3+7x^2+6x\)
Do đó \(x\left(x+1\right)\left(x+6\right)-x^3=\left(x^3+7x^2+6x\right)-x^3=7x^2+6x\)
\(\Rightarrow7x^2+6x=5x\Rightarrow7x^2=-x\Rightarrow7=\frac{-x}{x^2}=\frac{-x}{\left(-x\right).\left(-x\right)}=\frac{1}{-x}\)
\(\Rightarrow-x=\frac{1}{7}\Rightarrow x=-\frac{1}{7}\)
x(x + 1)(x + 6) - x^3 = 5x
(x^2 + x)(x + 6) - x^3 = 5x
x^3 + 6x^2 + x^2 + 6x - x^3 = 5x
(x^3 - x^3) + (6x^2 + x^2) + (6x - 5x) = 0
7x^2 + x = 0
x(7x + 1) = 0
TH1:
x = 0
TH2:
7x + 1 = 0
7x = -1
x = -1/7
Vậy x = 0 hoặc x = -1/7
