用因式分解解一元二次方程 4(x+3)*2=25(x-2)*2 (3x+2)(x-4)=0 9(2x+3)*2-4(2x-5)*2=0
4(x+3)*2=25(x-2)*2(3x+2)(x-4)=09(2x+3)*2-4(2x-5)*2=0...
4(x+3)*2=25(x-2)*2
(3x+2)(x-4)=0
9(2x+3)*2-4(2x-5)*2=0 展开
(3x+2)(x-4)=0
9(2x+3)*2-4(2x-5)*2=0 展开
1个回答
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4(x+3)*2=25(x-2)*2
(2x+6)^=(5X-10)^2
(2x+6)^-(5X-10)^2=0
(2X+6+5X-10)(2X+6-5X+10)=0
(7X-4)(16-3X)=0
X1=7/4 X2=16/3
(3x+2)(x-4)=0
X1=-3/2 X2=4
9(2x+3)^2-4(2x-5)^2=0
(6X+9)^2-(4X-10)^2=0
(6X+9+4X-10)(6X+9-4X+10)=0
(10X-1)(2X+19)=0
X1=1/10 X2=-19/2
(2x+6)^=(5X-10)^2
(2x+6)^-(5X-10)^2=0
(2X+6+5X-10)(2X+6-5X+10)=0
(7X-4)(16-3X)=0
X1=7/4 X2=16/3
(3x+2)(x-4)=0
X1=-3/2 X2=4
9(2x+3)^2-4(2x-5)^2=0
(6X+9)^2-(4X-10)^2=0
(6X+9+4X-10)(6X+9-4X+10)=0
(10X-1)(2X+19)=0
X1=1/10 X2=-19/2
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