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\(A=\frac{2015+2016+2017}{2014+2015+2016+2017+2018}x1000\)
a ) 1 + 2 + 3 + 4 + ... + x = 1275 ( có x số tự nhiên )
( x + 1 ) . x : 2 = 1275
( x + 1 ) . x = 1275 x 2
( x + 1 ) . x = 2550
( x + 1 ) . x = 50 . 51
Mà x , x + 1 là hai số tự nhiên liên tiếp => x = 50
Vậy x = 50
1+2+3+4+...+x=1275
\(\frac{x.\left(x+1\right)}{2}=1275\)
x(x+1)=1275x2=2550
x(x+1)=50.51
x=50
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Leftrightarrow x=\frac{2010}{10}=201\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Rightarrow x=\frac{2010}{10}=201\)
\(=\frac{1}{10}\)
ko ghi lại đề
ta thấy : 2019 - 1 = 2018
2020 - 2 = 2018
2021 - 3 = 2018
2022 - 4 = 2018
=> x = 2018
thử lại :
2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022
= 1 + 1 = 1 + 1
2 = 2
\(1\frac{2}{3}+x=2\frac{1}{3}\)
\(\frac{5}{3}+x=\frac{7}{3}\)
\(x=\frac{7}{3}-\frac{5}{3}\)
\(x=\frac{2}{3}\)
\(\frac{5}{3}+x=\frac{7}{3}\)
\(x=\frac{7}{3}-\frac{5}{3}\)
\(x=\frac{2}{3}\)
\(x+1\frac{1}{2}=\frac{3}{2}+\frac{1}{2}\)
\(x+1\frac{1}{2}=2\)
\(x=2-1\frac{1}{2}\)
\(x=\frac{1}{2}\)
\(X+\frac{1}{2}X=3\)
\(1X+\frac{1}{2}X=3\)
\(\left(1+\frac{1}{2}\right)X=3\)
\(\frac{3}{2}X=3\)
\(X=3\div\frac{3}{2}=2\)