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\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
c) (x+1) + (x+2) + ... + (x+5) = 90
=> 5x + ( 1 + 2 + ... + 5 ) = 90
5x + 15 = 90
5x = 90 - 15
5x = 75
x = 75 : 5
x = 15
d) (x+1) + (x+2) + .... + (x+100) = 20150
=> 100x + ( 1+2+...+100 ) = 20150
100x + 5050 = 20150
100x = 20150 - 5050
100x = 15100
x = 15100 : 100
x = 151
Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90
<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90
<=> 5x + 15 = 90
=> 5x = 75
=> x = 15
\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]:\left(11+x\right)=3\)
\(\left[\left(35\frac{5}{7}-5\frac{5}{7}\right)+2\frac{3}{4}+\frac{1}{4}\right]:\left(11+x\right)=3\)
\(\left[30+3\right]:\left(11+x\right)=3\)
\(33:\left(11+x\right)=3\)
\(11+x=11\)
\(x=0\)
\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\left(\frac{35\times7+5}{7}+\frac{2\times4+3}{4}\right)-\frac{5\times7+5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\left(\frac{250}{7}+\frac{11}{4}\right)-\frac{40}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\frac{1077}{28}-\frac{167}{28}\right]\div\left(11+x\right)=3\)
\(32,5\div\left(11+x\right)=3\)
\(11+x=32,5\div3\)
\(11+x=\frac{65}{6}\)
\(x=\frac{65}{6}-11=-\frac{1}{6}\)
\(A=\frac{2015+2016+2017}{2014+2015+2016+2017+2018}x1000\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Leftrightarrow x=\frac{2010}{10}=201\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Rightarrow x=\frac{2010}{10}=201\)
\(=\frac{1}{10}\)
(x+1) + (x+2)+(x+3)+(x+4)+(x+5)=55
=> 5x + 15=55
=>5x=55-15=40
=>x=40/5
Vây x= 8
a) 3 x (x + 1) + 4 x (x + 2) = 18
<=> 3x + 3 + 4x + 8 = 18
<=> 7x + 11 = 18
<=> 7x = 18 - 11
<=> 7x = 7
<=> x = 7 : 7 = 1
=> x = 1
b) 4 x (x + 1) + 7 x (2x + 3) = 61
<=> 4x + 4 + 14x + 21 = 61
<=> 18x + 25 = 61
<=> 18x = 61 - 25
<=> 18x = 36
<=> x = 36 : 18
<=> x = 2
=> x = 2