Rút gọn B=\(\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\cdot\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\)
K
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3 tháng 2 2022
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
L
29 tháng 5 2019
\(\left(\frac{1}{2+2.\sqrt{a}}+\frac{1}{2-2.\sqrt{a}}-\frac{a^2+1}{1-a^2}\right).\left(1+\frac{1}{a}\right)\)
\(=\left(\frac{2-2.\sqrt{a}+2+2.\sqrt{a}}{\left(2+2.\sqrt{a}\right)\left(2-2.\sqrt{a}\right)}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)\)
\(=\left(\frac{4}{4-4a}-\frac{a^2+1}{\left(1-a\right).\left(1+a\right)}\right).\left(\frac{a+1}{a}\right)=\frac{\left(1+a\right)}{\left(1-a\right).\left(1+a\right)}\cdot\frac{a+1}{a}=\frac{1+a}{\left(1-a\right).a}=\frac{a+1}{a-a^2}\)
16 tháng 3 2020
\(\text{GIẢI :}\)
ĐKXĐ : \(a\ne\pm1\).
\(M=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a^3-a}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\left(\frac{a}{a^2-1}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\left(\frac{a^2}{a\left(a^2-1\right)}-\frac{1}{a\left(a^2-1\right)}\right):\frac{a^2-2a+1}{a+a^3}\)
\(=\frac{1}{a^2-2a+1}-\frac{a^2-1}{a\left(a^2-1\right)}:\frac{\left(a-1\right)^2}{a\left(1+a^2\right)}\)
\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{a\left(a^2-1\right)}\cdot\frac{a\left(a^2+1\right)}{1+a^2}\)
\(=\frac{1}{a^2-2a+1}-\frac{\left(a-1\right)^2}{1+a^2}=\frac{-a^2}{\left(a-1\right)^2}\).
TH
25 tháng 2 2020
\(A=\left(\frac{1-a^3}{a-a^2}+1\right)\cdot\left(\frac{1+a^3}{1+a}-a\right):\frac{\left(1-a^2\right)^3}{1+a}\)
\(=\left(\frac{\left(1-a\right)\cdot\left(1+a+a^2\right)}{a\cdot\left(1-a\right)}+1\right)\cdot\left(\frac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right)\)\(:\frac{\left(1-a\right)^3\cdot\left(1+a\right)^3}{1+a}\)
\(=\left(\frac{1+a+a^2+a}{a}\right)\cdot\left(1-a+a^2-a\right):\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\frac{1+2a+a^2}{a}\cdot\left(1-2a+a^2\right):\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\frac{\left(1+a\right)^2}{a}\cdot\left(1-a\right)^2:\left[\left(1-a\right)^3\cdot\left(1+a\right)^2\right]\)
\(=\text{[}\frac{\left(1+a\right)^2}{a}:\left(1+a\right)^2\text{]}\cdot\text{[}\left(1-a\right)^2:\left(1-a\right)^3\text{]}\)
\(=\frac{1}{a}\cdot\frac{1}{1-a}=\frac{1}{a\left(1-a\right)}=\frac{1}{a-a^2}\)
Để \(A>A^2\Rightarrow\frac{1}{a-a^2}>\frac{1}{\left(a-a^2\right)^2}\)
Có ĐKXĐ : \(\left(a-a^2\right)\ne0\)
Mà \(\left(a-a^2\right)< \left(a-a^2\right)^2\)trừ trường hợp \(\left(a-a^2\right)=1\)
Từ tất cả điều trên suy ra : \(A\)thuộc tất cả các giá trị khác 1 để \(A>A^2\)
DP
1
KN
4 tháng 11 2019
\(M=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}-\frac{2a}{a^3-a^2+a-1}\right)\)
\(=\left(\frac{a^2+1}{a^2+1}+\frac{a}{a^2+1}\right):\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right)\)
\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{\left(a^2+1\right)\left(a-1\right)}\right)\)
\(=\frac{a^2+a+1}{a^2+1}:\frac{a-1}{a^2+1}=\frac{a^2+a+1}{a-1}\)
Trả lời
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\) \(\left(a\ge0.a\ne1\right)\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{1}{\left(a+1\right)^2}-\frac{1}{\left(a-1\right).\left(a+1\right)}\right]\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{a-1-a-1}{\left(a+1\right)^2.\left(a-1\right)}\right]\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.0\)
\(B=\frac{1}{a+1}\)
Vậy \(B=\frac{1}{a+1}\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)ĐK\left(a\ge0;a\ne1\right)\)
\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}-\frac{a^2+1}{\left(a^2-1\right)\left(a^2+1\right)}\right)\)
\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1-a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}\right)\)
\(=\frac{1}{a+1}\)
Vậy \(B=\frac{1}{a+1}\)