定义:经过有理式的恒等变形,任何有理式总能化为某个既约分式.如果这个既约分式是只含有一个自变数的真分式,还可进一步化为若干个既约真分式之和.这几个分式便称为原来那个既约分式的部分分式.
1.原式=[1-(t-3)^2/(t-1)^2]/[1+(t-1)^2/(t+1)^2]
=[4(t-2)/(t-1)^2]/[2(t^2+1)/(t+1)^2]
=[2(t-2)(t+1)^2]/[(t^2+1)(t-1)^2]
(本题原式的两个分母极可能都是t+1或t-1)
2.(1)原式=(x^2+2)/x(x+1)(x-2)=a/(x+1)+b/x+c/(x-2)
同乘(x+1),令x=-1,得a=1;
同乘x,令x=0,得b=-1;
同乘(x-2),令x=2,得c=1;
所以原式=1/(x+1)-1/x+1/(x-2)
(2)原式=(7x-3)/(x-1)(x+3)=a/(x-1)+b/(x+3)
同乘(x-1),令x=1,得a=1;
同乘x+3,令x=-3,得b=6.
所以原式=1/(x-1)+6/(x+3)
3.(1)[3(x-1)+2]/(x-1)=3+2/(x-1);
(2)[4-3(x+1)]/(x+1)=4/(x+1)-3;
(3)[(2/3)(3x+1)-1-2/3]/(3x+1)=2/3-(5/3)/(3x+1)
(4)[(x^4+x^2)+(x^2+1)+1]/(x^2+1)=x^2+1+1/(x^2+1)
(5)[(x^4+x^3-6x^2)+(x^3+x^2-6x)+1]/(x+3)(x-2)
=x^2+x+1/(x+3)(x-2)=x^2+x+1/5[1/(x-2)-1/(x+3)]
4.(题目有误,先用下面来解)
(5x-8)/(x^2-4x+4)=[5(x-2)+2]/(x-2)^2=5/(x-2)+2/(x-2)^2
太累了,希望采纳!