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共4个回答
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胡椒面瓜儿
丨Lv 3
将原方程移项,得1/(x+4)+1/(x+6)=1/(x+3)+1/(x+7) ==> (2x+10)/(x^2+10x+21)=(2x+10)/(x^2+10x+24) ==> (2x+10)[(x^2+10x+24)-(x^2+10x+21)]=0 ==> 6(x+5)=0 ==> x=-5。经检验(分式方程必须检验)知x=-5是原分式方程的解。
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lucky宝儿
丨Lv 3
(1/x+4)-(1/x+7)=(1/x+3)-(1/x+6 ) (1/x+4)+(1/x+6)=(1/x+3)+(1/x+7 ) [2x+10]/[x^+10x+24]=[2x+10]/[x^+10x+21](2x+10)[1/(x^+10x+24)-1/(x^+10x+21)]=01/(x^+10x+24)-1/(x^+10x+21)不等于0 2x+10=0x=-5经过验证x=-5是方程的根
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小小西北浪
丨Lv 3
题错了吧?括号没加!
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巧笑倩兮儿
丨Lv 3
题错了1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6 ) (x+7-x-4)/[(x+4)*(x+7)]=(x+6-x-3)/[(x+3)*(x+6)]3/[(x+4)*(x+7)]=3/[(x+3)*(x+6)](x+4)*(x+7)=(x+3)*(x+6)x^2+11x+28=x^2+9x+18x=-5
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