Sửa đề: 2020/1+2019/2+...+1/2020

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}}{\left(1+\dfrac{2019}{2}\right)+\left(1+\dfrac{2018}{3}\right)+...+\dfrac{1}{2020}+1+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}}{\dfrac{2021}{2}+\dfrac{2021}{3}+...+\dfrac{2021}{2020}+\dfrac{2021}{2021}}\)

=1/2021

1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)

2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)

3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)

4) \(ĐK:a< 0\)

5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)

`1/5 . 4/7 + 3/7 . 1/5 -1/5`

`=1/5 . 4/7 + 3/7 . 1/5 -1/5 . 1`

`=1/5 . ( 4/7+3/7-1)`

`=1/5 . ( 7/7-1)`

`= 1/5 . 0`

`=0`

\(\dfrac{1}{5}\times\dfrac{4}{7}+\dfrac{3}{7}\times\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right)=\dfrac{1}{5}\times0=0\)

1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)

2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)

3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)

4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)

5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)

6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)

A= 1/3 + 1/3^2 + ... + 1/3^8

3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)

3A=1+ 1/3 + 1/3^2+ ... +1/3^7

=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)

=> 2A= 1 - 1/ 3^8

2A= 6560/6561

A= 6560/6561 : 2

A= 3280/6561

nè bạn

Tính P = 

P=1+1/3+1/6+1/10+…..+1/2021×2022÷2

P/2=1/2+1/6+1/12+1/20+…..+1/2021×2022

P/2=1/1×2+1/2×3+1/3×4+…….+1/2021×2022

P/2=1-1/2+1/2-1/3+1/3-1/4+….+1/2021-1/2022=1-1/2022=2021/2022

P=2021/1011

Chúc bn học tốt

`A=1/(1.2)+1/(2.3)+1/(3.4)+....+1/(49.50)`

`=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50`

`=1-1/50=49/50`

hè công nhận ít câu hỏi thiệt !!

\(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)

\(=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+...+\dfrac{1}{\left(n-1\right)\cdot\left(n+1\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{\left(n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{3}{4}-\dfrac{1}{2n+2}\)

\(\lim\limits u_n=\lim\limits\left(\dfrac{3}{4}-\dfrac{1}{2n+2}\right)\)

\(=\lim\limits\dfrac{3}{4}-\lim\limits\dfrac{1}{2n+2}\)

\(=\dfrac{3}{4}-\lim\limits\dfrac{\dfrac{1}{n}}{2+\dfrac{1}{n}}\)

=3/4

=>Chọn A