Đáp án:
a,(x+2)²-5x=x(x-3)
⇔x²+4x+4-5x=x²-3x
⇔(x²-x²)+(4x-5x+3x)+4=0
⇔2x+4=0
⇔2x=-4
⇔x=-2
vậy S={-2}
b,
9x²-7x=0
⇔x(9x-7)=0
⇔ x=0
9x-7=0
⇔x=0
x=7/9
vậy S={0;7/9}
c, -x/x²-4 + x-4/x(x+2) = 5/x(x-2) (ĐK : x khác 0 , 2, -2 )
⇔x.(-x)/x(x-2)(x+2) + (x-4)(x-2)/x(x-2)(x+2) = 5(x+2)/x(x-2)(x+2)
⇒x² + (x-4)(x-2) = 5x+ 10
⇔x² + x² - 2x - 4x +8 -5x -10 = 0
⇔(x²-2x+1)+(x²-4x+4)-5x-10+3=0
⇔(x-1)²+(x-2)²-5x-7=0
ta có (x-1)²≥0 Vx
(x-2)²≥0 Vx
⇔(x-1)²+(x-2)²-5x≥0 Vx
⇔(x-1)²+(x-2)²-5x-7≥ -7 Vx
dấu "=" xảy ra ⇔ x-1=0 ⇔x= 1( TM)
x-2=0⇔x=2 (loại )
vậy S={1}
