作业帮这两道解方程题怎么做?
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这两道解方程题怎么做?
这两道解方程题怎么做?
这两道解方程题怎么做?
1)
(x+2)/(x+1)=[(x+1)+1]/(x+1)=(x+1)/(x+1)+1/(x+1)=1+1/(x+1)
同理：(x+4)/(x+3)=1+1/(x+3)
(x+6)/(x+5)=1+1/(x+5)
(x+8)/(x+7)=1+1/(x+7)
∴原方程可化为：
1/（x+1）-1/（x+3）=1/（x+5）-1/（x+7）
2/（x+1)(x+3)=2/(x+5)(x+7)
x²+4x+3=x²+12x+35
x=-4
经检验,x=-4是原方程的解.
2）
移项：1/(x+5)-1/(x+6)=1/(x+7)-1/(x+8)
1/(x+5)(x+6)=1/(x+7)(x+8)
(x+5)(x+6)=(x+7)(x+8)
x²+11x+30=x²+15x+56
x=-13/2
经检验,x=-13/2是原方程组的解.
1.先通分，即[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)] = [(x+6)(x+7)-(x+5)(x+8)]/[(x+5)(x+7)]
一化简，得 2/[(x+1)(x+3)] = 2/[(x+5)(x+7)]，(x+1)(x+3)= (x+5)(x+7)
x^2+4x+3 = x^2+12x+35 8x = -32 x=-4
2, ...
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1.先通分，即[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)] = [(x+6)(x+7)-(x+5)(x+8)]/[(x+5)(x+7)]
一化简，得 2/[(x+1)(x+3)] = 2/[(x+5)(x+7)]，(x+1)(x+3)= (x+5)(x+7)
x^2+4x+3 = x^2+12x+35 8x = -32 x=-4
2, 1/(x+5) - 1/(x+6) = 1/(x+7) - 1/(x+8)
两边各通分，得 1/[(x+5)(x+6)] = 1/[(x+7)(x+8)]
(x+5)(x+6) = (x+7)(x+8)
x^2+11x+30 = x^2+15x+56 4x=-26 x=-13/2
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