Đáp án:
10/ 9x²+12x+3=0
⇔(9x²+9x)+(3x+3)=0
⇔9x(x+1)+3(x+1)=0
⇔3(x+1)(3x+1)=0
⇔\(\left[ \begin{array}{l}x+1=0\\3x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=\frac{-1}{3}\end{array} \right.\)
11/ (x-1)³=0
⇔x-1=0
⇔x=1
12/ (x+1)³=27
⇔(x+1)³-3³=0
⇔ (x+1-3)[(x+1)²+3(x+1)+3²]=0
⇔(x-2)(x²+2x+1+3x+3+9)=0
⇔(x-2)(x²+5x+13)=0
⇔ x-2=0
⇔x=2
13/ (2x+3)³=-8
⇔(2x+3)³+2³=0
⇔(2x+3+2)[(2x+3)²-2(2x+3)+2²]=0
⇔(2x+5)(4x²+12x+9-4x-6+4)=0
⇔(2x+5)(4x²+8x+7)=0
⇔ 2x+5 =0
⇔ 2x=-5
⇔$\frac{-5}{2}$
14/ x³+3x²+3x+1=8
⇔(x+1)³=8
⇔(x+1)³-2³=0
⇔(x+1-2)[(x+1)²+2(x+1)+2²]=0
⇔(x-1)(x²+2x+1+2x+2+4)=0
⇔(x-1)(x²+4x+7)=0
⇔ x-1=0
⇔ x=1
15/ x³-3x²+3x-9=0
⇔(x³-3x²+3x-1)-8=0
⇔(x-1)³-2³=0
⇔(x-1-2)[(x-1)²+2(x-1)+2²]=0
⇔(x-3)(x²-2x+1+2x-2+4)=0
⇔(x-3)(x²+3)=0
⇔ x-3=0
⇔ x=3
16/ x³+3x²+3x=0
⇔x(x²+3x+3)=0
⇔ x=0
