(x-6)³=(x-6)²

=> (x-6) \(\in\){0;1}

=> x-6=0              hoặc x-6=1

=> x=0+6             hoặc x=1+6

=> x=6                 hoặc x=7

=> x \(\in\){6; 7}

\(a,\dfrac{3}{8}=\dfrac{6}{x}\\ \Rightarrow x=6:\dfrac{3}{8}\\ \Rightarrow x=16\\ b,\dfrac{1}{9}=\dfrac{x}{27}\\ \Rightarrow x=\dfrac{1}{9}.27\\ \Rightarrow x=3\\ c,\dfrac{4}{x}=\dfrac{8}{6}\\ \Rightarrow x=4:\dfrac{4}{3}\\ \Rightarrow x=3\\ d,\dfrac{3}{x-5}=\dfrac{-4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Rightarrow3x+6=-4x+20\\ \Rightarrow3x+6+4x-20=0\\ \Rightarrow7x-14=0\\ \Rightarrow7x=14\\ \Rightarrow x=2\)

a: =>6/x=3/8

hay x=16

b: =>x/27=1/9

nên x=3

c: =>4/x=4/3

nên x=3

d: =>3/x-5=-4/x+2

=>3x+2=-4x+20

=>7x=18

hay x=18/7

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

\(3\left(x-5\right)-2\left(2x-3\right)+3\left(6-x\right)=17\)

\(\Rightarrow3x-15-4x+6+18-3x=17\)

\(\Rightarrow-4x=8\)

\(\Rightarrow x=-2\)

Vậy x = -2

ghi đề khó hiểu quá 

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=0

7x^2+x=0

x(7x+1)=0

x=0 hoặc 7x+1=0<=> x=-1/7