(1/x^2+x-2)+(1/x^2+7x+10)=2求解,
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(1/x^2+x-2)+(1/x^2+7x+10)=2求解,
应该是 1/(x^2+x-2)+1/(x^2+7x+10)=2
==>1/((X+2)(X-1))+1/((X+2)(X+5))=2
==>(X+2)(X+5)+(X+2)(X-1)=2(X+2)(X-1)(X+2)(X+5)
==>X^2+7X+10+X^2+X-2=2(X+2)(X-1)(X+2)(X+5)
==>2(X^2+4X+4)=2(X+2)(X+2)=2(X+2)(X-1)(X+2)(X+5)
==>1=(X-1)(X+5)=X^2-4X-5 ==>X^2-4X+4-9=1
==>(X-2)^2=10 ==>X=+/-(根下10)+2
==>1/((X+2)(X-1))+1/((X+2)(X+5))=2
==>(X+2)(X+5)+(X+2)(X-1)=2(X+2)(X-1)(X+2)(X+5)
==>X^2+7X+10+X^2+X-2=2(X+2)(X-1)(X+2)(X+5)
==>2(X^2+4X+4)=2(X+2)(X+2)=2(X+2)(X-1)(X+2)(X+5)
==>1=(X-1)(X+5)=X^2-4X-5 ==>X^2-4X+4-9=1
==>(X-2)^2=10 ==>X=+/-(根下10)+2
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