Đáp án + Giải thích các bước giải:
`a//(3x-2)(4x+3)=(2-3x)(x-1)`
`⇔(3x-2)(4x+3)+(3x-2)(x-1)=0`
`⇔(3x-2)(4x+3+x-1)=0`
`⇔(3x-2)(5x+2)=0`
`⇔` \(\left[ \begin{array}{l}3x-2=0\\5x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\frac{2}{3}\\x=-\frac{2}{5}\end{array} \right.\)
Vậy `S={\frac{2}{3};-\frac{2}{5}}`
`b//x^{2}+(x+3)(5x-7)=9`
`⇔x^{2}+5x^{2}+8x-21=9`
`⇔6x^{2}+8x-30=0`
`⇔(6x^{2}+18x)-(10x+30)=0`
`⇔6x(x+3)-10(x+3)=0`
`⇔(x+3)(6x-10)=0`
`⇔` \(\left[ \begin{array}{l}x+3=0\\6x-10=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-3\\x=\frac{5}{3}\end{array} \right.\)
Vậy `S={-3;\frac{5}{3}}`
`c//2x^{2}+5x+3=0`
`⇔(2x^{2}+2x)+(3x+3)=0`
`⇔2x(x+1)+3(x+1)=0`
`⇔(x+1)(2x+3)=0`
`⇔` \(\left[ \begin{array}{l}x+1=0\\2x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1\\x=-\frac{3}{2}\end{array} \right.\)
Vậy `S={-1;-\frac{3}{2}}`
`d//` Câu `d` đề bài của bạn sai nhá !!!
- Mình sửa đề :
`(3-2x)/(3006)+(3-2x)/(2007)+(3-2x)/(2008)-(3-2x)/(2009)+(3-2x)/(2010)=0`
`⇔(3-2x)((1)/(3006)+(1)/(2007)+(1)/(2008)-(1)/(2009)+(1)/(2010))=0`
`⇔3-2x=0` . Vì `((1)/(3006)+(1)/(2007)+(1)/(2008)-(1)/(2009)+(1)/(2010)) \ne0`
`⇔2x=3`
`⇔x=\frac{3}{2}`
Vậy `S={\frac{3}{2}}`
