b, B = 1\(\frac{1}{1x2}+\frac{1}{2x3}+......+\frac{1}{99x100}\)
c, C = \(\frac{1}{1x2}+\frac{1}{2x3}+......+\frac{1}{n\left(n+1\right)}\)
d, D = 1 + 2 + 3 + ......+ n
Chứng minh: \(\frac{3}{\left(1x2\right)}+\frac{5}{\left(2x3\right)}+...+\frac{2n+1}{\left(n\left(n+1\right)\right)^2}=\frac{n\left(n+2\right)}{\left(n+1\right)^2}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+.....+\frac{1}{99x100}\)
Ta có :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\times2}+...+\frac{1}{99\times100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{99x100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
k cho mình nha bạn
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
tính nhanh
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+....+\frac{1}{99x100}\)
1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
tinh gia tri bieu thuc:
a) M = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{99x100}\)
b)N = \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{97x99}\)
\(x\)la dau nhan
gâp ạ
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(M=1-\frac{1}{100}\)
\(M=\frac{99}{100}\)
a)(864x11-423x4):(432x(3+6+9+...+27-132))
b)1x2+2x3+3x4+...+99x100
c)\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+...+\left(1+2+3+...+99\right)}{1x9+2x98+3x97\: +....+99x1}\)
giúp mk với , mk tick cho , nha !!!!
Mình chỉ tính câu b và c thội nhé!.
Ta có:
b) \(1.2+2.3+3.4+...+99.100\)
\(=\frac{99.100.101}{3}=333300\)
c) \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+99.1}\)
\(=\frac{1+1+2+1+2+3+1+2+3+4+...+1+2+3+...+99}{1.99+2.98+3.97+...+99.1}\)
\(=\frac{\left(1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+3+...+3\right)+....+99}{1.99+2.98+3.97+...+99.1}\)
\(=\frac{1.99+2.98+3.97+...+99.1}{1.99+2.98+3.97+...+99.1}=1\)
Tìm x trong biểu thức sau:
\(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{8x9}+\frac{1}{9x10}\right)x100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
Tìm số S
\(S=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{99x100}\)
Ghi cách giải ra nha!
\(\text{S}\)= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ .... + \(\frac{1}{99}\)- \(\frac{1}{100}\)
\(S\)= ( 1 - \(\frac{1}{100}\)) : 2
\(S\)= \(\frac{99}{100}\): 2
\(S\)= \(\frac{99}{200}\)
tick nhé Lê Thiên Hương
Tính bằng cách thuận tiện :
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{99x100}\)
Mình sẽ tick cho một bạn nhanh nhất ! ^_^
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Đặt \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(M=1-\frac{1}{100}\)
\(M=\frac{99}{100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+....+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)