Đáp án:
`a)S={9/20}`
`b)S={0;-5/2}`
`c)S={0}`
`d)S={6;-9}`
Giải thích các bước giải:
Câu 1:
`a)2x-2/5=1/2`
`⇔2x=1/2+2/5`
`⇔2x=5/10+4/10`
`⇔2x=9/10`
`⇔x=9/10:2`
`⇔x=9/10 . 1/2`
`⇔x=9/20`
Vậy `S={9/20}`
`b)(3x-7)(2x+5)-(2x+5)(2x-7)=0`
`⇔(2x+5)[(3x-7)-(2x-7)]=0`
`⇔(2x+5)(3x-7-2x+7)=0`
`⇔(2x+5)x=0`
`⇔`$\left[\begin{matrix} x=0\\ 2x+5=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0\\ x=-\dfrac{5}{2}\end{matrix}\right.$
Vậy `S={0;-5/2}`
`c)(x+2)/(x-2)-(x-2)/(x+2)=(4x²)/(x²-4)(ĐKXĐ:x`$\neq$ `±2)`
`⇔(x+2)/(x-2)-(x-2)/(x+2)=(4x²)/[(x+2)(x-2)]`
`⇔[(x+2)²]/[(x+2)(x-2)]-[(x-2)²]/[(x+2)(x-2)]=(4x²)/[(x+2)(x-2)]`
`⇒(x+2)²-(x-2)²=4x²`
`⇔x²+4x+4-(x²-4x+4)=4x²`
`⇔x²+4x+4-x²+4x-4=4x²`
`⇔x²+4x+4-x²+4x-4-4x²=0`
`⇔(x²-x²-4x²)+(4x+4x)+(4-4)=0`
`⇔-4x²+8x=0`
`⇔-4x(x-2)=0`
`⇔`$\left[\begin{matrix} -4x=0\\ x-2=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0(TM ĐKXĐ)\\ x=2(Ko TM ĐKXĐ)\end{matrix}\right.$
Vậy `S={0}`
`d)|2x+3|=15`
`⇔`$\left[\begin{matrix} 2x+3=15\\ 2x+3=-15\end{matrix}\right.$
`⇔`$\left[\begin{matrix} 2x=12\\ 2x=-18\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=6\\ x=-9\end{matrix}\right.$
Vậy `S={6;-9}`
