Mọi người giúp mình giải bài tập toán với ạ

1. Tìm x biết:

Mọi người giải giúp mình 1 đến 22 thôi ạ, Cảm ơn mọi người rất nhiều

Đáp án:

`1)4x(x+1)=8(x+1)`

`<=>4x(x+1)-8(x+1)=0`

`<=>(x+1)(4x-8)=0`

`<=>`$\left[\begin{matrix} x+1=0\\ 4x-8=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-1\\ x=2\end{matrix}\right.$`

`Vậy S={-1;2}`

`2)5x(x-2)-(2-x)=0`

` <=>5x(x-2)+(x-2)=0`

`<=>(x-2)(5x+1)=0`

`<=>`$\left[\begin{matrix} x-2=0\\ 5x+1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=2\\ x=\dfrac{-1}{5}\end{matrix}\right.$

`Vậy S={2;-1/5}`

`3)x(x-1)-2(1-x)=0`

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

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

`<=>`$\left[\begin{matrix} x-1=0\\ x+2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.$

`Vậy S={1;-2}`

`4)x(x+2)-3(x+2)=0`

`<=>(x+2)(x-3)=0`

`<=>`$\left[\begin{matrix} x+2=0\\ x-3=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-2\\ x=3\end{matrix}\right.$

`Vậy S={-2;3}`

`5)2x(x+1)-3(x+1)=0`

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

`<=>`$\left[\begin{matrix} x+1=0\\ 2x-3=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-1\\ x=\dfrac{3}{2}\end{matrix}\right.$

`Vậy S={-1;3/2}`

`6)x^2(x-2)+3(x-2)=0`

`<=>(x-2)(x^2+3)=0`

Vì `x^2>=0`

`=>x^2+3>0`

`<=>x-2=0`

`<=>x=2`

`Vậy S={2}`

`7)2x(x-5)-7(5-x)=0`

`<=>2x(x-5)+7(x-5)=0`

`<=>(x-5)(2x+7)=0`

`<=>`$\left[\begin{matrix} x-5=0\\ 2x+7=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=5\\ x=\dfrac{-7}{2}\end{matrix}\right.$

`Vậy S={5;-7/2}`

`8)3x(x-2)+2(x-2)=0`

`<=>(x-2)(3x+2)=0`

`<=>`$\left[\begin{matrix} x-2=0\\ 3x+2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=2\\ x=\dfrac{-2}{3}\end{matrix}\right.$

`Vậy S={2;-2/3}`

`9)x(3x+2)-x(2x+3)=0`

`<=>3x^2+2x-2x^2-3x=0`

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

`<=>x(x-1)=0`

`<=>`$\left[\begin{matrix} x=0\\ x-1=0\Rightarrow x=1\end{matrix}\right.$

`Vậy S={0;1}`

`10)5x(x-1)=x-1`

`<=>5x(x-1)-(x-1)=0`

`<=>(x-1)(5x-1)=0`

`<=>`$\left[\begin{matrix} x-1=0\\ 5x-1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=1\\ x=\dfrac{1}{5}\end{matrix}\right.$

`Vây S={1;1/5}`

`11)x(x+2)-x-2=0`

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

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

`<=>`$\left[\begin{matrix} x+2=0\\ x-1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-2\\ x=1\end{matrix}\right.$

`Vậy S={-2;1}`

`12)x(x-2)+x-2=0`

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

`<=>(x+2)(x+1)=0`

`<=>`$\left[\begin{matrix} x+2=0\\ x+1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-2\\ x=-1\end{matrix}\right.$

`Vậy S={-2;-1}`

`13)x(x-3)-x+3=0`

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

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

`<=>`$\left[\begin{matrix} x+3=0\\ x-1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-3\\ x=1\end{matrix}\right.$

`Vậy S={-3;1}`

`14)2x(x-1)-x+1=0`

`<=>2x(x-1)-(x-1)=0`

`<=>(x-1)(2x-1)=0`

`<=>`$\left[\begin{matrix} x-1=0\\ 2x-1=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=1\\ x=\dfrac{1}{2}\end{matrix}\right.$

`Vậy S={1;1/2}`

`15)x(x+3)-2x-6=0`

`<=>x(x+3)-2(x+3)=0`

`<=>(x+3)(x-2)=0`

`<=>`$\left[\begin{matrix} x+3=0\\ x-2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=-3\\ x=2\end{matrix}\right.$

`Vậy S={-3;2}`

`@Zenitsu`

Giải thích các bước giải: