1/(X∧2-10X-29)+1/(X∧2-10X-45)=2/(X∧2-10X-69)” 请问怎么解

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1/(X∧2-10X-29)+1/(X∧2-10X-45)=2/(X∧2-10X-69)” 请问怎么解
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1/(X∧2-10X-29)+1/(X∧2-10X-45)=2/(X∧2-10X-69)” 请问怎么解
1/(X∧2-10X-29)+1/(X∧2-10X-45)=2/(X∧2-10X-69)” 请问怎么解

1/(X∧2-10X-29)+1/(X∧2-10X-45)=2/(X∧2-10X-69)” 请问怎么解
设X^2-10X-29=t 则1/t+1/(t-16)=2/(t-40) 同乘后得到t=10
所以X^2-10X-29=10 得X=13或-3

设x^2-10x-69=y,得:1/(y+40)+1/(y+24)=2/y;2(y+32)/[(y+40)(y+24)]=2/y,y(y+32)=(y+40)(y+24),解得y=-30,所以x^2-10x-69=-30;解得x1=-3,x2=13,经检验...