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Ta có \(\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
\(=\frac{a^3+2a^2+2a+1-2a-2}{a^3+2a^2+2a+1}\)
\(=\frac{a^3+2a^2+2a+1}{a^3+2a^2+2a+1}-\frac{2a-2}{a^3+2a^2+2a+1}\)
\(=1-\frac{2a-1}{a^3+2a^2+2a+1}\)
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Giải:Ta có:\(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
\(=>A=\frac{\left(a+1\right).\left(a^2+a-1\right)}{\left(a+1\right).\left(a^2+a+1\right)}\)
\(=>A=\frac{a^2+a-1}{a^2+a+1}\)
\(=>A=\frac{-1}{1}\)
tk gium minh nha neu thay dung nha!
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Đặt biểu thức là A.
Ta có:
\(\frac{\left(a^3+a^2\right)+\left(a^2+1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)}=\frac{a^2\left(a+1\right)+\left(a+1\right)\left(a+1\right)}{a^2\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right)\left(a^2+a-1\right)}{\left(a+1\right)\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a+1}\).
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\(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}=\frac{a^3+a^2+a^2-1}{a^3+a^2+a^2+a+a+1}=\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}\)
\(A=\frac{a^2\left(a+1\right)+\left(a-1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right).\left(a^2+a-1\right)}{\left(a+1\right).\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a+1}\)
Vậy \(A=\frac{a^2+a-1}{a^2+a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(\frac{a^3+a^2+a^2-1}{a^3+a^2+a^2+a+a+1}=\frac{a^2\left(a+1\right)+\left(a-1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right)\left(a^2+a-1\right)}{\left(a+1\right)\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}=\frac{a^3+a^2+a^2-1}{a^3+a^2+a^2+a+a+1}=\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}\)
\(A=\frac{a^2\left(a+1\right)+\left(a-1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right).\left(a^2+a-1\right)}{\left(a+1\right).\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a+1}\)
Vậy A=..................
A=\(\frac{a^3+2a^2-1}{a^3+2a^2+2a+1}\)
A=\(\frac{a^3+2a^2+1-2}{a^3+2a^2+1+2a^2}\)
A=\(\frac{a^3+2a^2+1}{a^3+2a^2+1}+\frac{-2}{a^3+2a^2+1+2a^2}\)
A=\(1+\frac{-2}{a^3+2a^2+1+2a^2}\)
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 = \(\frac{a^2+a-1}{a^2+a+1}\)