a) ( 1+1/9) x ( 1+1/10 ) x ( 1+1/11) x .... x ( 1+1/120)
b) ( 1-1/10) x ( 1-1/11) x ( 1-1/12 ) x .... x ( 1-1/99)
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VM
1
AP
14 tháng 7 2020
\(A=\frac{9}{10}\times\frac{10}{11}\times\frac{11}{12}\times...\times\frac{99}{100}\)
\(A=\frac{9\times10\times11\times...99}{10\times11\times12\times...\times100}\)
\(\Rightarrow A=\frac{9}{100}\)
VM
0
VM
0
T
2
10 tháng 7 2020
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)
\(A=\left(-\frac{9}{10}\right)\cdot\left(-\frac{10}{11}\right)\cdot\left(-\frac{11}{12}\right)\cdot....\cdot\left(-\frac{98}{99}\right)\left(-\frac{99}{100}\right)\)
\(A=\frac{\left(-9\right)\left(-10\right)\left(-11\right)...\left(-98\right)\left(-99\right)}{10\cdot11\cdot12\cdot....\cdot99\cdot100}\)
\(A=\frac{9\cdot10\cdot11\cdot....\cdot98\cdot99}{10\cdot11\cdot12\cdot...\cdot99\cdot100}=\frac{9}{100}\)
T
1
10 tháng 7 2020
Bài làm:
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)
\(A=\frac{\left(-9\right)}{10}.\frac{\left(-10\right)}{11}.\frac{\left(-11\right)}{12}...\frac{\left(-98\right)}{99}.\frac{\left(-99\right)}{100}\)
\(A=-\left(\frac{9.10.11...98.99}{10.11.12...99.100}\right)\)
\(A=-\frac{9}{100}\)
Học tốt!!!!
TS
1
10 tháng 4 2017
\(\frac{1}{9x10}\)\(+\frac{1}{10x11}\)\(+\frac{1}{11x12}\)\(+.....\)\(+\frac{1}{805x806}\)
\(=\frac{1}{9}\)\(-\frac{1}{10}\)\(+\frac{1}{10}\)\(-\frac{1}{11}\)\(+\frac{1}{11}\)\(-\frac{1}{12}\)\(+.....\frac{1}{805}\)\(-\frac{1}{806}\)
\(=\frac{1}{9}\)\(-\frac{1}{806}\)
\(=\frac{797}{7254}\)
CN
1
NH
20 tháng 6 2019
a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Đặt:
\(X=\left(1+\dfrac{1}{9}\right)\left(1+\dfrac{1}{10}\right)\left(1+\dfrac{1}{11}\right).....\left(1+\dfrac{1}{200}\right)\)
\(X=\dfrac{10}{9}.\dfrac{11}{10}.\dfrac{12}{11}......\dfrac{201}{200}\)
\(X=\dfrac{10.11.12......201}{9.10.11......200}\)
\(X=\dfrac{201}{9}\)
\(Y=\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{12}\right).....\left(1-\dfrac{1}{99}\right)\)
\(Y=\dfrac{9}{10}.\dfrac{10}{11}.\dfrac{11}{12}.....\dfrac{98}{99}\)
\(Y=\dfrac{9.10.11......98}{10.11.12.....99}\)
\(Y=\dfrac{9}{99}=\dfrac{1}{11}\)
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