Rút gọn biểu thức sau: T=(\(\frac{1}{2}\)+1)×(\(\frac{1}{3}\)+1)×(\(\frac{1}{4}\)+1)×……………×(\(\frac{1}{98}\)+1)×(\(\frac{1}{99}\)+1)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta xét biểu thức sau :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left[\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\right]}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(với n > 0)
Áp dụng : \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+...+\left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right)\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}=\sqrt{100}-1=10-1=9\)
A = \(\frac{1}{1+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + . . . . . . . . . + \(\frac{1}{\sqrt{99+\sqrt{100}}}\)
= \(\sqrt{2}\) - 1 + \(\sqrt{2}\) - \(\sqrt{3}\) + . . . . . . . + \(\sqrt{100}\) - \(\sqrt{99}\)
= - 1 + \(\sqrt{100}\) = \(\sqrt{100}\) - 1 = 10 - 1 = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 99/1+98/2+97/3+...+1/99=(98/2+1)+(97/3+1)+...+(1/99+1)+1
=100/2+100/3+...+100/99+100/100
=100(1/2+1/3=1/4+1/5+...+1/99+1/100)
Vậy (1/2+1/3+...+1/100)/((99/1+98/2+...+1/99)=1/100
![](https://rs.olm.vn/images/avt/0.png?1311)
xét mẫu số = \(\frac{99}{1}\)+\(\frac{98}{2}\)+....+\(\frac{1}{99}\)
mẫu số = (\(1+\frac{98}{2}\))+(\(1+\frac{97}{3}\))+.......+(\(1+\frac{1}{99}\))
mẫu số = \(\frac{100}{2}\)+\(\frac{100}{3}\)+....+\(\frac{100}{99}\)
mẫu số =100 x (\(\frac{1}{2}\)+\(\frac{1}{3}\)+....+\(\frac{1}{99}\)) (1)
thay (1) vào biểu thức trên
1/2+1/3+1/4+.....+1/100 / 100 x (1/2+1/3+...+1/99)
= \(\frac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~
![](https://rs.olm.vn/images/avt/0.png?1311)
sao lại lấy ảnh của tui.
bài cậu hỏi tôi làm rồi đó
nhớ ****
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}\)
\(A=\left(\frac{2.3....99}{1.2....98}\right).\left(\frac{2.3....99}{3.4....100}\right)\)
\(A=\frac{99}{1}.\frac{2}{100}\)
\(A=\frac{198}{100}\)
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).......\left(\frac{1}{98}+1\right).\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right).....\left(\frac{1}{98}+\frac{98}{98}\right).\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(T=\frac{3.4.5....99.100}{2.3.4.....98.99}\)
\(T=\frac{100}{2}\)
\(T=50\)
Vậy T = 50
Chúc bạn học tốt!
dua tao chich roi tao tra loi