Đáp án:
Giải thích các bước giải:
Bài 1:
`c)5(x-3)(x-2)-3(x-5)(x-2)=0`
`↔(5x-15)(x-2)+(-3x+15)(x-2)=0`
`↔(x-2)(5x-15-3x+15)=0`
`↔2x(x-2)=0`
`↔` \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy phương trình có tập nghiệm `S={0;2}`
`e)(x-5/3)(x+3)=0`
`↔` \(\left[ \begin{array}{l}x-\dfrac{5}{3}=0\\x+3=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=-3\end{array} \right.\)
Vậy `S={5/3;-3}`
`g)x(x^2-1)=0`
`↔x(x-1)(x+1)=0`
`↔` \(\left[ \begin{array}{l}x=0\\x-1=0\\x+1=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy `S={0;1;-1}`
`h)(x^2-2x+1)-4=0`
`↔(x-1)^2-2^2=0`
`↔(x-1-2)(x-1+2)=0`
`↔(x-3)(x+1)=0`
`↔` \(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
Vậy `S={3;-1}`
