Giải:

Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)

\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{5}+\dfrac{1}{13}\cdot3+\dfrac{1}{61}\cdot3\\ =\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{3}{60}=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

=> Điều phải chứng minh

Ta có :

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét :

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

bài giải:

đặt biểu thức bằng A

=> A= \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

ta thấy:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{13}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)

=> A<\(\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}\)<\(\dfrac{1}{2}\)

=> đpcm.

Lời giải:

Ta có:

\(\left\{\begin{matrix} \frac{1}{13}< \frac{1}{12}\\ \frac{1}{14}< \frac{1}{12}\\ \frac{1}{15}< \frac{1}{12}\end{matrix}\right.\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}(1)\)

\(\left\{\begin{matrix} \frac{1}{61}< \frac{1}{60}\\ \frac{1}{62}< \frac{1}{60}\\ \frac{1}{63}< \frac{1}{60}\end{matrix}\right.\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)

Hay \( \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)

Ta có đpcm.

Đặt A là biểu thức đó

Ta có:

\(\dfrac{1}{13}< \dfrac{1}{12};\dfrac{1}{14}< \dfrac{1}{12};\dfrac{1}{15}< \dfrac{1}{12}\)

\(\Rightarrow\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}\)

Ta cũng có

\(\dfrac{1}{61}< \dfrac{1}{60};\dfrac{1}{62}< \dfrac{1}{60};\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

\(\Rightarrow\)dpcm

a) Giải:

Ta có: \(4n-5=4\left(n-3\right)+7\)

Để \(\left(4n-5\right)⋮\left(n-3\right)\Leftrightarrow7⋮n-3\)

\(\Rightarrow n-3\inƯ\left(7\right)\)

Mà \(Ư\left(7\right)\in\left\{\pm1;\pm7\right\}\)

Nên ta có bảng sau:

Vậy \(n=\left\{2;4;-4;10\right\}\)

b) Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)

Giải:

Ta có: \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)\) \(+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Dễ thấy:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S< \dfrac{1}{2}\) (Đpcm)

Từ (1), (2), (3) suy ra:

bài 2

Hơi nhầm xíu 113 . 7^2+8^2=113 cứ tưởng 112. Hơi ngáo tí =[[

Lời giải

Biến đổi tương đương ta được: \(L=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}=\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+\dfrac{1}{4^2+5^2}+\dfrac{1}{5^2+6^2}+\dfrac{1}{6^2+7^2}+\dfrac{1}{7^2+8^2}\)

\(L=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+...+\dfrac{1}{7^2+\left(7+1\right)^2}\)

Chứng minh 1 bđt cơ bản sau: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\) thật vậy:

\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1=2n\left(n+1\right)+1>2n\left(n+1\right)\)

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

trở lại bài toán ta có: \(L< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+...+\dfrac{1}{2.7.8}\)

\(L< \dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..+\dfrac{1}{7.8}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{8}\right)=\dfrac{1}{2}-\dfrac{1}{16}< \dfrac{1}{2}\left(đpcm\right)\)

Đề sai đúng hk? CHỗ kia 112 chứ lấy đâu ra 113

p/s : 7^2+8^2=112. =))

Nếu cứ khăng khăng là 113 đúng thì bạn hãy cứ giữ suy nghĩ đó đi.

Đề bài tổng quát:

\(B=\dfrac{1}{n_1^2+\left(n_1+1\right)^2}+\dfrac{1}{n^2_2+\left(n_2+1\right)^2}+...+\dfrac{1}{n^2_n+\left(n_n+1\right)^2}\)

2 số chính phương liên tiếp cộng với nhau không có thể là 113. T tính ra 112 nên nghĩ you viết nhầm thôi.

Nếu k tin thì tối đi học xem đề thế nào nhé,t khẳng định luôn,đề này là 112