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Ta có: \(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}=\frac{\left(a^3+a^2-a\right)+\left(a^2+a-1\right)}{\left(a^3+a^2+a\right)+\left(a^2+a-1\right)}=\frac{a\left(a^2+a-1\right)+\left(a^2+a-1\right)}{a\left(a^2+a+1\right)+\left(a^2+a+1\right)}=\frac{\left(a+1\right)\left(a^2+a-1\right)}{\left(a+1\right)\left(a^2+a+1\right)}=\frac{\left(a^2+a-1\right)}{\left(a^2+a+1\right)}\)
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Ta có \(\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
\(=\frac{a^3+2a^2+2a+1-2a-2}{a^3+2a^2+2a+1}\)
\(=\frac{a^3+2a^2+2a+1}{a^3+2a^2+2a+1}-\frac{2a-2}{a^3+2a^2+2a+1}\)
\(=1-\frac{2a-1}{a^3+2a^2+2a+1}\)
A=1+3+32+33+34+....+311
3A=3(1+3+32+33+34+....+311)
3A=3+32+33+34+35+....+312
3A-A=(3+32+33+34+35+....+312)-(1+3+32+33+34+....+311)
2A=312-1
A=312-1+3
A=312+2
A=531443
Ta có:
A = \(\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
A = \(\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+1\right)+\left(2a^2+2a\right)}\)
A = \(\frac{a^2\left(a+1\right)+\left(a-1\right)\left(a+1\right)}{\left(a+1\right)\left(a^2-a+1\right)+2a\left(a+1\right)}\)
A = \(\frac{\left(a^2+a-1\right)\left(a+1\right)}{\left(a+1\right)\left(a^2-a+1+2a\right)}\)
A = \(\frac{a^2+a-1}{a^2+a+1}\)
\(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}=\frac{a^3+a^2+a^2-1}{a^3+a^2+a^2+a+a+1}=\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}\)
\(A=\frac{a^2\left(a+1\right)+\left(a-1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right).\left(a^2+a-1\right)}{\left(a+1\right).\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a+1}\)
Vậy \(A=\frac{a^2+a-1}{a^2+a+1}\)