cần số kg đường để trộn hết 10kg dâu là :

                   \(10.4:6=\dfrac{20}{3}\left(kg\right)\)

                          \(ds...\)

Đề yêu cầu chứng tỏ \(3^{n+2}-2^{n+4}+3^n+2^n⋮30\forall n\) nguyên dương à bạn?

\(3^{n+2}-2^{n+4}+3^n+2^n\)

\(=3^n.9-2^n.16+3^n+2^n\)

\(=\left(3^n.9+3^n\right)+\left(2^n-2^n.16\right)\)

\(=3^n.10-15.2^n\)

Ta có:

\(\left\{{}\begin{matrix}3^n⋮3\\10⋮10\end{matrix}\right.\Rightarrow3^n.10⋮30\) (1)

\(\left\{{}\begin{matrix}15⋮15\\2^n⋮2\end{matrix}\right.\Rightarrow15.2^n⋮30\) (2)

Từ (1) và (2) \(\Rightarrow3^n.10-15.2^n⋮30\)

\(\Rightarrowđpcm.\)

\(1+1=3\)

Ta thấy: B là tích của 99 số âm

\(\Rightarrow B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{100^2}\right)\)

\(=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}...\dfrac{9999}{10^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{99.101}{100^2}\)

\(=\dfrac{1.2.3...98.99}{2.3.4...99.100}.\dfrac{3.4.5...100.101}{2.3.4...99.100}\)

\(=\dfrac{1}{2}.\dfrac{101}{100}\)

\(=\dfrac{101}{200}>\dfrac{1}{2}\)

\(\Rightarrow B< -\dfrac{1}{2}\).

ủa sao từ \(\dfrac{1}{2^2}-1\) lại thành \(1-\dfrac{1}{2^2}\) vậy bạn

Ta có:

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...

\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)

\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)

\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)

\(\Rightarrow P< 1\)

Em chú ý mẫu

2 = 1 x 2

6 = 2 x 3

......

110 = 10 x 11

Từ đó em đưa về dạng tổng quát \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)

Rút gọn đi ta sẽ có được đáp án.

Chúc em học tốt!

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{10.11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}\)

\(=\dfrac{10}{11}\).

Ta có:

\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{10}\)

...

\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>100.\dfrac{1}{10}=10\).

`1+1xx10xx(-795)+1`

`=1+10xx(-795)+1`

`=1+(-7950)+1`

`=-7950+2`

`=-7948`

'0'

'''0'''

chu vi hay diện tích 2 tam giác bằng nhau