Toán rút gọn phân thức

V

vipboycodon

b) $\dfrac{1}{a^2+a}+\dfrac{1}{a^2+3a+2}+\dfrac{1}{a^2+5a+6}+\dfrac{1}{a^2+7a+12}+\dfrac{1}{a^2+9a+20}$
= $\dfrac{1}{a(a+1)}+\dfrac{1}{(a+1)(a+2)}+\dfrac{1}{(a+2)(a+3)}+\dfrac{1}{(a+3)(a+4)}+\dfrac{1}{(a+4)(a+5)}$
= $\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}$
= $\dfrac{1}{a}-\dfrac{1}{a+5}$
= $\dfrac{5}{a(a+5)}$
 
H

huuthuyenrop2

1/(a-b) + 1/(a+b) + 2a/(a^2+b^2) + 4a^3/(a^4+b^4) + 8a^7/(a^8+b^8)
$= \frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}\frac{8a^7}{a^8+b^8}$
Cứ như vậy cộng lại rồi rút gọn
 
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