Đề ???

a. 3x(x - 2) - x + 2 = 0

<=> 3x² - 6x - x + 2 = 0

<=> 3x² - 7x + 2 = 0

<=> 3x² - 6x - x + 2 = 0

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

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

<=> \(\left[{}\begin{matrix}3x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\end{matrix}\right.\)

b. x²(x + 1) + 2x(x + 1) = 0

<=> (x² + 2x)(x + 1) = 0

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

<=> \(\left[{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)

c. x(2x - 3) - 2(3 - 2x) = 0

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

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

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

đáp án thứ 2 nha

ghê

A có 100 phần tử

c. \(\dfrac{1}{2}x\left(x^2-4\right)+4\left(x+2\right)\)

= \(\dfrac{1}{2}x\left(x-2\right)\left(x+2\right)+4\left(x+2\right)\)

= \(\left[\dfrac{1}{2}x\left(x-2\right)-1\right]\left(x+2\right)\)

= \(\left(\dfrac{1}{2}x^2-1-1\right)\left(x+2\right)\)

= \(\left(\dfrac{1}{2}x^2-2\right)\left(x+2\right)\) 

a. A = (a + b)³ - (a - b)³

A = \(\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

A = (a + b - a + b)\(\left[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right]\)

A = 2b(a² + a² + a² + 2ab - 2ab + b² - b² + b²)

A = 2b(3a² + b²)

A = 6a²b + 2b³