3xx+x = 0

xx-5x-6 = -6

2xx+ 10x -12 = -12

xxx-xx+x -1 =-1

Nhanh lên nhé các bạn ai nhanh nhất mình sẽ cho 1 k

Giúp mình nha mình đang cần gấp

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

x(x+1)(x-6)-x³ = 5x

 <=>(x²+x)(x-6)-x³=5x

<=>x³-6x²+x²-6x-x³=5x

<=>-5x²-6x=5x

<=>-5x²-6x-5x=0

<=>-5x²-11x=0

<=>-x.(5x-11)=0

<=>x=0 hoặc 5x-11=0

<=>x=0 hoặc x=11/5

a) Tương đương      b) Không tương đương.