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解分式方程(初二)
(1)(2x²+1)\(x+2)=2x
(2)(x-2)\(x+2)-1=3\(x²-4)
(3) 3\(x-2 )+x\(2-x) = -2
人气:346 ℃ 时间:2020-05-21 06:58:55
解答

(2x²+1)\(x+2)=2x
(2x²+1)\(x+2)-2x=0
(2x²+1)\(x+2)-2x (x+2)\ (x+2)=0
(2x²+1-2x²-4x)/(x+2)=0
(1-4x)/(x+2)=0
1-4x=0
x=1/4

(x-2)\(x+2)-1=3\(x²-4)
((x-2)(x-2))/((x+2)(x-2))-1=3/(x²-4)
(x²-4x+4)/(x²-4)-1=3/(x²-4)
(x²-4x+4)-(x²-4)=3
x²-4x+4-x²+4=3
-4x+8=3
-4x=-5
x=5/4

3\(x-2 )+x\(2-x) = -2
3/(x-2)+x/-(x-2)=-2
3/(x-2)-x/(x-2)=-2
(3-x)/(x-2)=-2
3-x=-2(x-2)
3-x=-2x+4
x=1
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