Question

Tìm x :

a/2x(x - 1) - x(4 - x) = 0

b/3(1 - 4x)(x - 1) + 4(3x - 2)(x + 3)=-27

Answer 1

Giải:

a) \(2x\left(x-1\right)-x\left(4-x\right)=0\)

\(\Leftrightarrow2x^2-2x-4x+x^2=0\)

\(\Leftrightarrow3x^2-6x=0\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

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

Vậy ...

b) \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)

\(\Leftrightarrow\left(3-12x\right)\left(x-1\right)+\left(12x-8\right)\left(x+3\right)=-27\)

\(\Leftrightarrow3x-12x^2-3+12x+12x^2-8x+36x-24=-27\)

\(\Leftrightarrow43x-27=-27\)

\(\Leftrightarrow43x=0\)

\(\Leftrightarrow x=0\)

Vậy ...

Answer 2

a) 2x(x-1)-x(4-x)= 0

2x² - 2x - 4x + x²= 0

3x² - 6x = 0

3x(x-2) = 0

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b) \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=27\)

\(3\left(5x-4x^2-1\right)+4\left(3x^2+7x-6\right)-27=0\)

\(15x-12x^2-3+12x^2+28x-24-27=0\)

\(43x=0\)

\(x=0\)