Đáp án:
Phía dưới
Giải thích các bước giải:
`1)`
`a)2x(1-x)+5=9-2x^2`
`⇔2x-2x^2+5-9+2x^2=0`
`⇔2x-4=0`
`⇔x=2`
Vậy `x=2`
`b) 3x^3-14x^2-5x=0`
`⇔x(3x^2-14x-5)=0`
`⇔x(3x^2+x-15x-5)=0`
`⇔x[x(3x+1)-5(3x+1)=0`
`⇔x(x-5)(3x+1)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-5=0\\3x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=5\\x=-\dfrac{1}{3}\end{array} \right.\)
Vậy `x=0` hoặc `x=5` hoặc `x=-1/3`
`2)`
`x^2(x-2)+(4x-3)(2-x)`
`=x^2(x-2)-(4x-3)(x-2)`
`=(x^2-4x+3)(x-2)`
`=(x^2-3x-x+3)(x-2)`
`=[x(x-3)-(x-3)](x-2)`
`=(x-1)(x-3)(x-2)`
