Đáp án + Giải thích các bước giải:
Bài 1:
`a)`
`3x²(2x²-5x-4)`
`= 3x².2x²-3x².5x-3x².4`
`= 6x^4-15x³-12x²`
`b)`
`(x+1)²+ (x-2)(x+3)-4x`
`= x²+2x+1+x²+3x-2x-6-4x`
`= (x²+x²)+(2x+3x-2x-4x)+(1-6)`
`= 2x²-x-5`
Bài 2:
`a) 7x²+14xy= 7x(x+2y)`
`b) 3(x+4)-x²-4x`
`= 3(x+4)-x(x+4)`
`= (x+4)(3-x)`
`c) x²-2xy+y²-z²`
`= (x-y)²-z²`
`= (x-y-z)(x-y+z)`
`d) x²-2x-15`
`= x²-5x+3x-15`
`= x(x-5)+3(x-5)`
`= (x-5)(x+3)`
Bài 3:
`a) 7x²+2x=0`
`⇔ x(7x+2)=0`
`⇔` $\left[\begin{matrix} x=0\\ 7x+2=0\end{matrix}\right.$ `⇔`$\left[\begin{matrix} x=0\\ x=\dfrac{-2}{7}\end{matrix}\right.$
Vậy `x=0` hoặc `x=(-2)/7`
`b) x(x+4)-x²-6x=10`
`⇔ x²+4x-x²-6x=10`
`⇔ -2x=10`
`⇔ x=-5`
Vậy `x=-5`
`c) x(x-1)+2x-2=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 `x=1` hoặc `x=-2`
`d) (3x-1)²-(x+5)²=0`
`⇔ (3x-1-x-5)(3x-1+x+5)=0`
`⇔ (2x-6)(4x+4)=0`
`⇔` $\left[\begin{matrix} 2x-6=0\\ 4x+4=0\end{matrix}\right.$`⇔` $\left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.$
Vậy `x=3` hoặc ` x=-1`
