(2-x)^3+(2+x)^3-12x(x+1)=0

=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)

=>\(12x^2+16-12x^2-12x=0\)

=>16-12x=0

=>4-3x=0

=>x=4/3

`a)5x(x-1)-(x+2)(5x-7)=6`

`<=>5x^2-5x-(5x^2-7x+10x-14)=6`

`<=>5x^2-5x-(5x^2+3x-14)=6`

`<=>-8x+14=6`

`<=>8x=8<=>x=1`

Vậy `x=1`

`b)(x+2)^2-(x^2-4)=0`

`<=>x^2+4x+4-x^2+4=0`

`<=>4x+8=0`

`<=>4x=-8`

`<=>x=-2`

Vậy `x=-2`

a)x=5/2

b)x=-2

a)5x.(x-1)-(x+2).(5x-7)=6

<=> 5x²-5x-(5x²-7x+10x-14)=6

<=>  5x²-5x-5x²+7x-10x+14=6

<=> -8x+14=6

<=> -8x=-8 => x=1

Vậy x=1

b) (x+2)²-(x²-4)=0

<=> (x+2)²-(x²-2²)=0 <=> (x+2)²-(x-2)(x+2)=0

<=> (x+2)[(x+2)-(x-2)]=0

<=> (x+2)(x+2-x+2)=0

<=> (x+2).4=0

=> x+2=0

=> x=-2

Vậy x=-2

a) \(\left(5-x\right)\left(x-1\right)=-2x\left(x-1\right)\)

\(\Rightarrow\left(5-x\right)\left(x-1\right)+2x\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(5-x+2x\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

b) \(\left(x+3\right)^2-\left(x-13\right)\left(x+13\right)=0\)

\(\Rightarrow x^2+6x+9-x^2+169=0\)

\(\Rightarrow6x=-178\Rightarrow x=-\dfrac{89}{3}\)

a: =>2x^2=4

=>x^2=2

=>\(x=\pm\sqrt{2}\)

b: =>(x+1)^2-4=0

=>(x+1+2)(x+1-2)=0

=>(x+3)(x-1)=0

=>x=1 hoặc x=-3

c: =>(2x-1)^2-3^2=0

=>(2x-1-3)(2x-1+3)=0

=>(2x-4)(2x+2)=0

=>x=2 hoặc x=-1

d: x^2-x=0

=>x(x-1)=0

=>x=0 hoặc x=1

a: \(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a) \(x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)

e) \(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)

h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

f)\(=>x\left(2+1\right)=0=>3x=0=>x=0\)

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

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

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