xét ΔABC vuông tại A có \(cosB=\dfrac{AB}{BC}\)

=>\(\dfrac{6}{BC}=cos30=\dfrac{\sqrt{3}}{2}\)

=>\(BC=6\cdot\dfrac{2}{\sqrt{3}}=4\sqrt{3}\left(cm\right)\)

\(\left|A+B\right|< =\left|A\right|+\left|B\right|\)

=>\(\left(\left|A+B\right|\right)^2< =\left(\left|A\right|+\left|B\right|\right)^2\)

=>\(A^2+B^2+2AB< =A^2+B^2+2\left|AB\right|\)

=>2AB<=2|AB|

=>AB<=|AB|(luôn đúng)

Dấu '=' xảy ra khi AB>=0

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)

=>\(\dfrac{6}{BC}=sin30=\dfrac{1}{2}\)

=>\(BC=6\cdot2=12\left(cm\right)\)

Tỉ số giữa thời gian hoàn thiện một chiếc khẩu trang sau khi cải tiến và trước khi cải tiến là:

\(\dfrac{100\%}{100\%+60\%}=\dfrac{1}{1,6}=\dfrac{5}{8}\)

=>Phần trăm thời gian hoàn thiện giảm đi là:

\(1-\dfrac{5}{8}=\dfrac{3}{8}=37,5\%\)

Ta có:

\(290=2\cdot5\cdot29\\ 895=5\cdot197\\ 578=2\cdot17^2\\ =>ƯCLN\left(290;895;578\right)=1\)

\(290=2.5.29\)

\(895=5.179\)

\(578=2.17^2\)

Nên không tồn tại \(UCLN\left(290;895;578\right)\)

Ta có: `997.1001 `

`= (999 - 2) . (999 + 2) `

`= 999 . 999 - 2 . 999 + 2 . 999 - 4`

`= 999 . 999 - 4 < 999 . 999`

Vậy  `997.1001  < 999 . 999`

\(999.999=\left(1000-1\right)\left(1000-1\right)=\left(1000-1\right)^2=1000^2-2.1000+1\)

\(997.1001=\left(1000-3\right)\cdot\left(1000+1\right)=1000^2-2.1000-2\)

mà \(1>-2\Rightarrow1000^2-2.1000+1>1000^2-2.1000-2\)

\(\Rightarrow999.999>997.1001\)

S = 1 x 2+2 x 3+3 x 4+...+25 x 26

3S = 1 x 2 x 3 +2 x 3 x (4 - 1) +3 x 4 x (5 - 2) +...+25 x 26 x (27 - 24)

3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 25 x 26 x 27 - 24 x 25 x 26

3S = 25 x 26 x 27

3S = 17550

S = 5850

1234567890^2

1.524157875x10¹⁸

Đặt \(A=1\cdot2023+2\cdot2022+3\cdot2021+\cdots+2023\cdot1\)

\(=2\cdot\left(1\cdot2023+2\cdot2022+3\cdot2021+\cdots+1011\cdot1013\right)+1012^2\)

\(=2\cdot\left\lbrack1\left(2024-1\right)+2\left(2024-2\right)+\cdots+1011\left(2024-1011\right)\right\rbrack+1012^2\)

\(=2\cdot\left\lbrack2024\left(1+2+\cdots+1011\right)-\left(1^2+2^2+\cdots+1011^2\right)\right\rbrack+1012^2\)

\(=2\cdot\left\lbrack2024\cdot1011\cdot\frac{1012}{2}-\frac{1011\left(1011+1\right)\left(2\cdot1011+1\right)}{6}\right\rbrack+1012^2\)

\(=2\cdot\left\lbrack1011\cdot1012^2-1011\cdot1012\cdot\frac{2023}{6}\right\rbrack+1012^2\)

\(=2\cdot\left\lbrack1011\cdot1012^2-337\cdot506\cdot2023\right\rbrack+1012^2\)

\(=1012^2\cdot2022-337\cdot1012\cdot2023+1012^2\)

\(=1012^2\cdot2023-337\cdot1012\cdot2023\)

\(=1012\cdot2023\left(1012-337\right)=1012\cdot2023\cdot675\)

Đặt \(B=1\cdot2+2\cdot3+\cdots+2022\cdot2023\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+2022\left(2022+1\right)\)

\(=\left(1^2+2^2+\cdots+2022^2\right)+\left(1+2+\cdots+2022\right)\)

\(=\frac{2022\left(2022+1\right)\left(2\cdot2022+1\right)}{6}+\frac{2022\cdot2023}{2}\)

\(=337\cdot2023\cdot4045+1011\cdot2023\)

\(=337\cdot2023\left(4045+3\right)=337\cdot2023\cdot4048\)

Ta có: \(\frac{1\cdot2023+2\cdot2022+3\cdot2021+\cdots+2023\cdot1}{1\cdot2+2\cdot3+\cdots+2022\cdot2023}\)

\(=\frac{1012\cdot2023\cdot675}{337\cdot2023\cdot4048}=\frac{1012}{4048}\cdot\frac{675}{337}=\frac14\cdot\frac{675}{337}=\frac{675}{1348}\)