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2 tháng 8 2017

A=7+7+73+...+7n-1+7n

7A = 72 + 73 + 74 + ... + 7n + 7n+1

7A - A = ( 72 + 73 + 74 + ... + 7n + 7n+1 ) - ( 7+7+73+...+7n-1+7n )

6A = 7n+1 - 7

A = \(\frac{7^{n+1}-7}{6}\)

\(\dfrac{7}{8}+\dfrac{7}{120}+\dfrac{7}{360}+\dfrac{7}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)

\(=\dfrac{7}{1\cdot8}+\dfrac{7}{8\cdot15}+\dfrac{7}{360}+\dfrac{1}{7n-6}-\dfrac{1}{7n+1}+\dfrac{1}{7n+1}\)

\(=1-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{7}{360}+\dfrac{1}{7n-6}\)

\(=\dfrac{14}{15}+\dfrac{7}{360}+\dfrac{1}{7n-6}=\dfrac{343}{360}+\dfrac{1}{7n-6}\)

\(=\dfrac{343\left(7n-6\right)+360}{360\left(7n-6\right)}\)

\(=\dfrac{2401n-1698}{360\left(7n-6\right)}\)

14 tháng 1

\(\dfrac{7}{8}+\dfrac{7}{120}+\dfrac{7}{360}+\dfrac{7}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\\ =\left(\dfrac{7}{8}+\dfrac{7}{120}+\dfrac{7}{360}\right)+\left(\dfrac{7}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\right)\\ =\left(\dfrac{315}{360}+\dfrac{21}{360}+\dfrac{7}{360}\right)+\left(\dfrac{7}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{7n-6}{\left(7n+1\right)\left(7n-6\right)}\right)\)

\(=\dfrac{343}{360}+\dfrac{7n+1}{\left(7n-6\right)\left(7n+1\right)}\\ =\dfrac{343}{360}+\dfrac{1}{7n-6}\\ =\dfrac{343\left(7n-6\right)+360}{360\left(7n-6\right)}\\ =\dfrac{2401n-2058+360}{360\left(7n-6\right)}\\ =\dfrac{2401n-1698}{360\left(7n-6\right)}\)

 

NV
14 tháng 1

\(=\dfrac{7}{1.8}+\dfrac{7}{8.15}+\dfrac{7}{15.24}+...++\dfrac{7}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)

\(=1-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{24}+...+\dfrac{1}{7n-6}-\dfrac{1}{7n+1}+\dfrac{1}{7n+1}\)

\(=1\)

8 tháng 4 2018

Đặt \(A=\frac{4n+3}{7n+1}-\frac{3n-2}{7n+1}+\frac{2n-3}{7n+1}\) ta có : 

\(A=\frac{4n+3-3n+2+2n-3}{7n+1}\)

\(A=\frac{3n+2}{7n+1}\)

Vậy \(A=\frac{3n+2}{7n+1}\)

Chúc bạn học tốt ~ 

8 tháng 4 2018

(4n+3-(3n-2)+(2n-3))/7n+1

(4n+3-3n+2+2n-3)/7n+1

=(3n-2)/7n+1

6 tháng 5 2018

Ta có:

Đề kiểm tra 45 phút Đại số 11 Chương 4 có đáp án (Đề 4)

Chọn B.

29 tháng 3 2017

\(\frac{19n+7}{7n+11}=2\)

\(\Rightarrow x=3\)

Còn cách giải thì k xong mình nói

\(\left(\frac{-.-}{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{_{ }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}\right)\)

Có: `343=7^3`.

`A=7^n(1+7+49)=3.19.343`

`<=> 7^n.3.19=3.19.7^3`.

`<=> n=3.`

Vậy `n=3`

14 tháng 10 2023

7ⁿ + 7ⁿ⁺¹ + 7ⁿ⁺² = 3.19.343

7ⁿ.(1 + 7 + 7²) = 19551

7ⁿ.57 = 19551

7ⁿ = 19551 : 57

7ⁿ = 343

7ⁿ = 7³

n = 3

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

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

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

 

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

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

\(=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)