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theo công thức, ta tính đc:
A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51
=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)
\(=\frac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\)
\(=\frac{2\sqrt{a}}{\sqrt{a}}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(B=\frac{1}{a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{ĐKXĐ: }x\ne1\)
\(M=\frac{a^2+2}{a^3-1}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}=\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a+1}{a^2+a+1}-\frac{1}{a-1}\)
\(=\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a^2+2+a^2-1-a^2-a-1}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a^2-a}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{a.\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a}{a^2+a+1}\)
Với 0 < a < 1 ta có :
\(a+1>0\Rightarrow|a+1|=a+1\)
\(a-1< 0\Rightarrow|a-1|=-\left(a-1\right)=-a+1\)
Do đó ta có:
\(T=\frac{|a+1|+|a-1|}{|a+1|-|a-1|}=\frac{\left(a+1\right)-a+1}{\left(a+1\right)+a+1}=\frac{2}{2a}=\frac{1}{a}\)
Vậy \(T=\frac{1}{a}\)
@ Việt Hoàng @ chắc là em bị nhầm một chút chỗ này :)
\(T=\frac{\left(a+1\right)+\left(-a+1\right)}{\left(a+1\right)-\left(-a+1\right)}=\frac{a+1-a+1}{a+1+a-1}=\frac{2}{2a}=\frac{1}{a}\)