初三一元二次方程.求大家告我几道.(1) (x-2)^2=(2x+3)^2(2) (3x+2x)(x+3)=x+14(3) (y+3)(1-3y)=1+2y^2(4) (x-7)(x+3)+(x-1)(x+5)=38

来源:学生作业帮助网 编辑:作业帮 时间:2024/09/13 12:51:29
初三一元二次方程.求大家告我几道.(1) (x-2)^2=(2x+3)^2(2) (3x+2x)(x+3)=x+14(3) (y+3)(1-3y)=1+2y^2(4) (x-7)(x+3)+(x-1)(x+5)=38
x){1Ɏ';6?lgv>_ѭlc%˟tb׳Ol~O= @BȊ3**o\mTdem+ M@P`Qƕ yCmJ&p+̑  „LAz-lIX!L_5;aG AX0@lhna8XQkRmbkiZBf@kl  C[]C}cVF 1`m

初三一元二次方程.求大家告我几道.(1) (x-2)^2=(2x+3)^2(2) (3x+2x)(x+3)=x+14(3) (y+3)(1-3y)=1+2y^2(4) (x-7)(x+3)+(x-1)(x+5)=38
初三一元二次方程.求大家告我几道.
(1) (x-2)^2=(2x+3)^2
(2) (3x+2x)(x+3)=x+14
(3) (y+3)(1-3y)=1+2y^2
(4) (x-7)(x+3)+(x-1)(x+5)=38

初三一元二次方程.求大家告我几道.(1) (x-2)^2=(2x+3)^2(2) (3x+2x)(x+3)=x+14(3) (y+3)(1-3y)=1+2y^2(4) (x-7)(x+3)+(x-1)(x+5)=38
1)
(x-2)²=(2x+3)²
x²-4x+4=4x²+12x+9
3x²+16x+5=0
(3x+1)(x+5)=0
x1=-1/3
x2=-5