\(B=\frac{3.4+2.9.16+3.12.20+4.15-24}{4.5+2.12.20+3.16.20+4.20.30}\)
\(C=\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{3.2}-\frac{1}{2.1}\)
Dấu chấm là dấu x
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+\frac{1}{4.5}=x\)
Tìm x biết dấu chấm là dấu nhân.
\(x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(x=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
x=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5
x=1-1/5
x=4/5
4\(\frac{1}{2}\)-\(\frac{15}{10.9}-\frac{15}{9.8}-...-\frac{15}{3.2}-\frac{15}{2.1}\)
\(\frac{6}{2.5}+\frac{8}{5.9}+\frac{12}{9.15}\)
\(1-\frac{2}{2.1}-\frac{6}{2.5}-\frac{8}{5.9}-\frac{8}{9.13}\)
\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{2015.2016}\)Dấu chấm là nhân nha mấy bạn.
\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+....+\frac{3}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1}-\frac{1}{2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{2015}{2016}\)
\(\Rightarrow S=\frac{2015}{672}\)
Vậy: \(\Rightarrow S=\frac{2015}{672}\)
Bạn giải giúp mk câu mk đăng tầm 5 phút nha!
S=1/1-1/2+1/2-1/3+1/3-1/4+1\5-..............................-1/2015+1/2016
S=1/1-1/2016
S=2015/2016
Giúp mình với nhé:
Tính:
a)\(\left(\frac{1}{2^2-1}\right).\left(\frac{1}{3^2-1}\right).\left(\frac{1}{4^2-1}\right)...\left(\frac{1}{98^2-1}\right).\left(\frac{1}{99^2-1}\right)\)
b)\(4\frac{1}{2}-\frac{15}{10.9}-\frac{15}{9\cdot8}-\frac{15}{3.2}-\frac{15}{2.1}\)
\(1\frac{1}{9}\cdot1\frac{1}{10}\cdot1\frac{1}{11}\cdot...\cdot1\frac{1}{2011}\)
dấu chấm ở giữa kia là dấu nhân
1 1/9 x 1 1/10 x 1 1/11 x ... x 1 1/2011
=10/9 x 11/10 x 12/11 x ... x 2012/2011
khử
còn 2012/9
=\(\frac{10}{9}\)x\(\frac{11}{10}\)x\(\frac{12}{11}\)x.........x\(\frac{2012}{2011}\)
=\(\frac{2012}{9}\)
\(1\frac{1}{9}.1\frac{1}{10}.1\frac{1}{11}...1\frac{1}{2011}\)
= \(\frac{10}{9}.\frac{11}{10}.\frac{12}{11}...\frac{2012}{2011}\)
= \(\frac{2012}{9}\)
Bài 1 : tính tổng sau
P = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
S = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
Q = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
Dấu " . " là dấu nhân nhé
Bài 2 Tính tích P = ( \(1+\frac{1}{2}\)) X ( \(1+\frac{1}{3}\)) x ( \(1+\frac{1}{4}\)) x ... x ( \(1+\frac{1}{99}\))
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
Tính nhanh :
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}...-\frac{1}{3.2}-\frac{1}{2.1}\)
giúp mk với nha các bạn
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(A=\frac{1.1.2.2.3.3...9.9}{1.2.2.3.3.4...9.10}\)
\(A=\frac{1}{10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-1\right)\)
\(B=\frac{1}{99}-\frac{1}{99}+1\)
\(B=1\)
sorry nha Thiên Sứ đội lốt Ác Quỷ mk 5 - 6
Bài 9: tính
a, A= 1+2+3+4+....+100
b,B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
c, C=\(\frac{10}{56}+\frac{10}{140}+\frac{10}{200}+.......+\frac{1}{1400}\)
a) 1 + 2 + 3 + 4 + ... + 100
= (100 + 1) x 100 : 2
= 5050
a) A=(100-1):1+1=100 số hạng
A=100:2=50 cặp
tính giá trị của từng cặp số = (1+100)+(2+99)+(3+98)+...+(50+51)=101
tính giá trị của biểu thức A: 50*101=5050
[ mình tính theo công thức đó ]
Tính:
\(\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{10}-.-\frac{92}{100}\right):\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+.+\frac{1}{500}\right)\)
dấu\(.\)nghĩa là dấu 3 chấm nhé! ai xong và đúng sẽ có tích.
#)Giải :
\(\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{10}-...-\frac{92}{100}\right):\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right)\)
\(=\left(1-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+...+1-\frac{92}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)
\(=\left(\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)
\(=8\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)
\(=8\div\frac{1}{5}\)
\(=40\)
#~Will~be~Pens~#