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\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)
Vậy x = 117/50
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)
\(=\left(1-\frac{1}{10}\right).100\)
\(=\frac{9}{10}.100\)
= 90
Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)
\(\Rightarrow x+\frac{266}{100}=5\)
\(\Rightarrow x=\frac{117}{50}\)
Vậy \(x=\frac{117}{50}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{15.17}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{17}\)
\(A=1-\frac{1}{17}\)
\(A=\frac{16}{17}\)
\(B=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)
\(B=\frac{4}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(B=\frac{4}{2}\left(1-\frac{1}{13}\right)\)
\(B=\frac{4}{2}\cdot\frac{12}{13}\)
\(B=\frac{24}{13}\)
=> A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}\)
=> A= \(\frac{1}{1}-\frac{1}{17}\)
=> A= \(\frac{16}{17}\)
\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{13}\right)\)
\(\Rightarrow B=2.\frac{12}{13}\)
\(\Rightarrow B=\frac{24}{13}\)
bài A: áp dụng công thức: 1 + 2 + 3 + ... + n = n x (n + 1) : 2 tính được 5050
bài B: áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rồi triệt tiêu gần hết, qui đồng mẫu số tính được B = 99/100
A = 1 + 2 + 3 + 4 + 5 + ... + 99 + 100
= ( 100 + 1 ) x 100 : 2 = 5050
Vậy A = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy \(B=\frac{99}{100}\)
Học tốt #
\(\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{37.39}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{37}-\frac{1}{39}\)
\(=\frac{1}{3}-\frac{1}{39}\)
\(=\frac{13}{39}-\frac{1}{39}\)
\(=\frac{12}{39}=\frac{4}{13}\)
ta có A=1/3-1/5+1/5-1/7+1/7-1/9+....+1/37-1/39
=1/3-1/39
=12/39
\(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n-1}\)
Áp dụng ta có:
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Tính C tương tự, áp dụng:
\(\frac{2}{n\left(n+2\right)}=\frac{n+2-n}{n\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)
B = 9899/9900
C=I don't know !!
Ủng hộ nhé !
a,(11/15+4/15)+(5/7+2/7)
=1+1
=2
b,5/9x(1/2+6/4)
=5/9x2
=10/9
c,1/2:(7/8+9/8)
=1/2:2
=1
d,(17/10-7/10)+1/2
=1+1/2
=3/2
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=2\)
b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)
c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)
\(=2\div\frac{1}{2}=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}=\frac{3}{2}\)
a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)
\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)
\(=1+1\)
\(=2\)
b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)
\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)
\(=\frac{25}{81}.\frac{3}{4}\)
\(=\frac{25}{108}\)
c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)
\(=\frac{7}{8}.2+\frac{9}{8}.2\)
\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)
\(=2.\frac{16}{8}\)
\(=2.2\)
\(=4\)
d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)
\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)
\(=1+\frac{1}{2}\)
\(=\frac{2}{2}+\frac{1}{2}\)
\(=\frac{3}{2}\)
b) 2/3.5+2/5/7+2/7.9+2/9.11+...+2/13.15
=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+..+1/13-1/15
=1/3-1/15
=4/15
c) 2/1.2+2/2.3+2/3.4+..+2/8.9+2/9.10
=2(1/1.2+1/2.3+1/3.4+..+1/8.9+1/9.10)
=2(1-1/2+1/2-1/3+1/3-1/4+..+1/8-1/9+1/9-1/10)
=2(1-1/10)
=2.9/10=9/5
d) 1/3+1/9+1/27+...+1/729
đặt A=1/3^1+1/3^2+1/3^3+..+1/3^6
3A=1+1/3+1/3^2+...+1/3^5
3A-A=1+1/3+1/3^2+...+1/3^5-1/3-1/3^2-1/3^3-...-1/3^6
2A=1-1/3^6
2A=728/729
A=728/729:2
A=364/729