(20+2x)(40-x)=1200,怎样解,

(X+X*20%)-[(1200-X)-(1200-X)*20%]=40% X+X*20%-(1200-X)-(1200-X)*20%=40% (40-X)(20+2X)=1200 (20+2x)(40-x)=1200 （20+2x)(40-x)=1200 (40-x)(20+2x)=1200 (40-x)(20+2x)=1200 (20+2x)(40-x)=1200 (40-x)(20+2x)=1200 (40-x)(20+2x)=1200 800+80X-20X-2x^2=1200 -2x^2+60X-400=0 x^2-30X+200=0 (X-10)(X-20)=0 X-10=0 X-20=0(X-10)(X-20)=0 40-x)(20+2x)=1200 800+80X-20X-2x^2=1200 -2x^2+60X-400=0 ^2-30X+200=0 (X-10)(X-20)=0 60(30x)=40x+20(30+x)+1200 X-40%X+1200=2/3X怎么算 20/(x-20)=(2x-40)/(x+40) 2x/40%=x+20x=? X+(X+40)+(2X-20)=320 x+40=2x x=? 2(40-2x)(20-x)+2x(20-x)+2x(40-2x)=550怎么解 ((20＋2X)(40-x)==1200求解