\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)

\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)

\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)

\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)

\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)

\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)

\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)

\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)

\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)

Giải:

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)

\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)

\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)

\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)

Vậy \(A=\dfrac{30}{31}.\)

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}............................\dfrac{899}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}..............................\dfrac{29.31}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3.2.4.3.5..........29.31}{2.2.3.3.4.4.........30.30}\)

\(\Leftrightarrow A=\dfrac{\left(2.3.........29.30\right).\left(3.4.5......29.31\right)}{\left(2.3....29.30\right).\left(2.3.4.......29.30\right)}\)

\(\Leftrightarrow A=\dfrac{31}{2.30}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....................................\dfrac{2499}{2500}\)

\(\Leftrightarrow B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.............................\dfrac{49.51}{50.50}\)

\(\Leftrightarrow B=\dfrac{\left(2.3.4.....49\right).\left(4.5.6......51\right)}{\left(3.4.5....50\right)\left(3.4.5.....50\right)}=\dfrac{2.51}{50.3}=\dfrac{17}{25}\)

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

Từ đó ta có:

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{50^2-1}{50^2}>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+...+1-\dfrac{1}{49.50}\)

\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)

Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)

\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)

a) Ta có \(A=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)

\(=\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\dfrac{4\cdot6}{5\cdot5}\cdot...\cdot\dfrac{49\cdot51}{50\cdot50}\)

\(=\dfrac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot49\cdot51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)

\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\cdot\dfrac{4\cdot5\cdot6\cdot...\cdot51}{3\cdot4\cdot5\cdot...\cdot50}\)

= \(\dfrac{2}{50}\cdot17=\dfrac{17}{25}\)

b) Vì n nguyên nên 3n - 1 nguyên

Để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên thì 12 ⋮ ( 3n - 1 ) hay ( 3n - 1 ) ϵ Ư_{( 12 )}

Ư_{( 12 )} = { \(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\) }

Lập bảng giá trị 

Vì n nguyên nên n ϵ { 0; 1; -1 } 

Vậy n ϵ { 0; 1; -1 } để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên

a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)

\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)

Vậy \(A< \dfrac{1}{2}\)

b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)

49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)

b) Đặt :

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)

\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)

\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)

\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))

\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)

Lại có :

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)

Mà :

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}< 1\)

\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)

\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)

\(\Rightarrow A>48\) \(\rightarrowđpcm\)

Ta có:

\(C=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)

\(\Rightarrow C=1-\dfrac{1}{4}+1-\dfrac{1}{9}+...+1-\dfrac{1}{2500}\)

\(\Rightarrow C=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...+1-\dfrac{1}{50^2}\)

\(\Rightarrow C=\left(1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\) (có \(49\) chữ số \(1\))

\(\Rightarrow C=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)

Lại có:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

Mà \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}< 1\)

\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>-1\)

\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)>49-1=48\)

Vậy \(C=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}>48\) (Đpcm)

sai

= \(49-\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=\dfrac{4949}{102}\notin N\)

Vậy \(S\notin N\)

\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)\(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)

\(\Rightarrow S=1-\dfrac{1}{4}+1-\dfrac{1}{9}+1-\dfrac{1}{16}+...+1-\dfrac{1}{2500}\)

\(\Rightarrow S=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)

\(\Rightarrow S=\left(1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)

Từ 2-50 có 49 số nên có 49 số 1

\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< 49\)

Nhận xét: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{50}=1-\dfrac{1}{50}< 1\)

\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)>-1\)

\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)

\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>48\) (2)

Từ (1) và (2) \(\Rightarrow48< S< 49\)

Vậy \(S\notin N\)

A=2.4/3^2 . 3.5/4^2 . 4.6/5^2 ............ . 49.51/50^2

A=2/3-51/50

A=17/25.

Chúc bạn hok tốt.

Bài này cũng dễ ý mà, vô cùng đơn giản.........

Giải:

Ta có: \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}.\)

\(=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{49.51}{50^2}.\)

\(=\dfrac{\left(2.3.4.....49\right)\left(4.5.6.....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}.\)

\(=\dfrac{2.51}{3.50}.\)

\(=\dfrac{17}{25}.\)

CHÚC BN HỌC TỐT!!! ^ _ ^

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https://hoc24.vn/hoi-dap/question/214681.html

A= 3^2-1/3.3 . 4^2-1/4.4 . 5^2-1/5.5 . ... 50^2-1/50.50 A= (3+1).(3-1).(4+1).(4-1).(5+1).(5-1). ... (50+1).(50-1) / 3.3.4.4.5.5. ... . 50.50 A=4.2.5.3.6.4. ... 51.49 / 3.3.4.4.5.5....50.50 A=(4.5.6. ... .51).(2.3.4. ... 49)/(3.4.5.... .50).(3.4.5.. ... 50) A= 51.2/3.50 A=17/25

Ta có:

\(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}......\dfrac{2499}{2500}\)

= \(\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}......\dfrac{49.51}{50.50}\)

= \(\dfrac{2.4.3.5.4.6......49.51}{3.3.4.4.5.5......50.50}\)

= \(\dfrac{\left(2.3.4....49\right)\left(4.5.6....51\right)}{\left(3.4.5....50\right)\left(3.4.5....50\right)}\)

= \(\dfrac{2}{50}.\dfrac{51}{3}\) = \(\dfrac{17}{25}\)