Cho A=1/2.2/3.3/4. ... .79/89
CMR : A<1/9
Cho A=1/2.2/3.3/4. ... .79/89
CMR : A<1/9
Ai nhanh mình tick
Cho :A= \(\frac{1}{2.2}\) +\(\frac{1}{3.3}\) +\(\frac{1}{4.4}\)+....\(\frac{1}{1009.1009}\)
CMR A<\(\frac{3}{4}\)
Ta có:
\(\frac{1}{2.2}\)<\(\frac{1}{1.2}\)
\(\frac{1}{3.3}\)<\(\frac{1}{2.3}\)
..............
\(\frac{1}{1009.1009}\)<\(\frac{1}{1008.1009}\)
=>A< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1008}-\frac{1}{1009}\)
=\(\frac{1}{1}-\frac{1}{1009}=\frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)
=>A<\(\frac{3}{4}\)
Mình nghĩ bạn cần xem lại :
\(A< \frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)không có nghĩa là \(A< \frac{3}{4}\)
Xem lại ..
M=1/2.2/3.3/4.4/5.......99/100 .Cmr 1/15<M<1/110
Lời giải:
$M=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{99}{100}$
$=\frac{1.2.3.4....99}{2.3.4...100}=\frac{1}{100}$
Hiển nhiên $\frac{1}{15}> \frac{1}{100}> \frac{1}{110}$ nên ta có đpcm.
** Sửa đề: CMR: $\frac{1}{15}> M> \frac{1}{110}$
CMR: 1/2.2 +1/3.3+1/4.4+....+1/100.100<1
Mik cần gấp
Ta có : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}< 1\)(đpcm)
+)Ta thấy:\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
............................
..............................
\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...............+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+............+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...............+\frac{1}{100.100}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..............+\frac{1}{99}-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+.............+\frac{1}{100.100}< 1\left(\text{Đ}PCM\right)\)
Chúc bạn học tốt
A=1+2.2!+3.3!+4.4!+...+100.100!
Tính tổng A = 1! + 2.2! + 3.3! + ....+ 99.99!
Ta có: \(n.n!=\left(n+1-1\right).n!=\left(n+1\right)!-n!\)
Suy ra \(A=\left(2!-1!\right)+\left(3!-2!\right)+...+\left(100!-99!\right)\)\(=100!-1!\)
Vậy.... (chắc hết rút gọn được rồi nhỉ)
A=1!+2.2!+3.3!+...+100.100!
Nhanh giùm !!!
=2.1!-1!+3.2!-2!+4.3!-3!+...+101.100!-100!
=2!-1!+3!-2!+4!-3!+...+101!-100!
=101!-1
Gọi n! (n giai thừa)
n! = 1 . 2 . 3 . 4 . ..... . n
Tính: A = 1.1! + 2.2! + 3.3! + 4.4! + 5.5!
1.1!+2.2!+3.3!+4.4!+5.5!+2.1.2+3.1.2.3+4.1.2.3.4+5.1.2.3.4.5
=1+4+18+96+600=600+96+4+18+1=600+100+19+=719
A = 1+ 1/ 2.2 + 1 / 3.3 +1/ 4.4 + ....+1/99.99 + 1/ 100. 100