Đáp án:
a/ (x-3)²-(x-3)(3-x²)=0
⇔(x-3)(x-3-3+x²)=0
⇔(x-3)(x+x²)=0
⇔x(x-3)(x+1)=0
⇔\(\left[ \begin{array}{l}x=0\\x-3=0 ; x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=3 ; x=-1\end{array} \right.\)
b/ 4(x+1)²+(2x-1)²- 8(x-1)(x+1) = 11
⇔ 4(x²+2x+1)+(4x²-4x+1)-8(x²-1)=11
⇔4x²+8x+4+4x²-4x+1-8x²+8-11=0
⇔(4x²+4x²-8x²)+(8x-4x)+(4+1+8-11)=0
⇔4x+2=0
⇔2(2x+1)=0
⇒2x+1=0
⇔x=$\frac{-1}{2}$
c/ (x-3)²-(x+3)(x+5)=3x+4
⇔(x²-6x+9)-(x²+5x+3x+15)=3x+4
⇔x²-6x+9-x²-5x-3x-15-3x-4=0
⇔(x²-x²)+(-6x-5x-3x-3x)+(9-15-4)=0
⇔-17x-2=0
⇔-17x=2
⇔x=$\frac{-2}{17}$
d/ 3(x-5)²+2x(x-5)=0
⇔(x-5)[3(x-5)+2x]=0
⇔(x-5)(3x-15+2x)=0
⇔(x-5)(x-15)=0
⇔\(\left[ \begin{array}{l}x-5=0\\x-15=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=15\end{array} \right.\)
e/(x-5)²-x²+25=0
⇔(x-5)²-(x²-25)=0
⇔(x-5)²-(x-5)(x+5)=0
⇔(x-5)(x-5-x-5)=0
⇔-10(x-5)=0
⇒x-5=0
⇔x=5
f/ (2x-1)²-(x+3)²=0
⇔(2x-1-x-3)(2x-1+x+3)=0
⇔(x-4)(3x+2)=0
⇔\(\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=\frac{-2}{3}\end{array} \right.\)
g/ 4(2+3x)(3x-2) - (6x+1)²=7
⇔-4(2+3x)(2-3x)-(6x+1)²=7
⇔-4(4-9x²)-(36x²+12x+1)=7
⇔-16+36x²-36x²-12x-1=7
⇔(36x²-36x²)-12x-(1+16)=7
⇔-12x-17=7
⇔-12x=7+17
⇔-12x=24
⇔x=-2
h/ 3(2x-1)²-6x(2x-3)=6
⇔3(4x²-4x+1)-6x(2x-3)=6
⇔12x²-12x+3-12x²+18x=6
⇔(12x²-12x²)+(-12x+18x)=6-3
⇔6x=3
⇔x=$\frac{1}{2}$
