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11 tháng 2 2022

Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}\)

\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\)

Mặt khác:

\(\dfrac{7}{12}=\dfrac{20}{60}+\dfrac{20}{80}\)

mà \(\left\{{}\begin{matrix}\dfrac{20}{60}< \left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)\\\dfrac{20}{80}< \left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\end{matrix}\right.\)

⇒ \(\dfrac{7}{12}< A\) (1)

Ta có:

\(\dfrac{5}{6}=\dfrac{20}{40}+\dfrac{20}{60}\)

mà \(\left\{{}\begin{matrix}\dfrac{20}{40}>\left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)\\\dfrac{20}{60}>\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\end{matrix}\right.\)

⇒ \(A< \dfrac{5}{6}< 1\)(2)

Từ (1) và (2)

⇒ \(\dfrac{7}{12}< A< 1\) (đpcm)

24 tháng 3 2021

Ta có:

A=9999931999−5555571997

A=9999931998.999993−5555571996.555557

A=(9999932)999.999993 − (5555572)998.555557

A=\(\overline{\left(....9\right)}^{999}\) . 999993 - \(\overline{\left(...1\right)}.\text{555557}\)

A=\(\overline{\left(...7\right)}-\overline{\left(...7\right)}\)

A= \(\overline{\left(...0\right)}\)

Vì A có tận cùng là 0 nên \(A⋮5\)

4 tháng 5 2018

Ta có:

7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 => (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 và 1/61> 1/62> ... >1/79> 1/80 => (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 => 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12 => ĐPCM

19 tháng 8 2018

Ta có : 1/41 + 1/42 + ... + 1/60 > 1/60 * 20 = 1/3 .

1/61 + 1/62 + ... + 1/80 > 1/80 * 20 = 1/4 .

1/41 + 1/42 + ... + 1/80 > 1/3 + 1/4 = 4/12 + 3/12 .

= 7/12 .

Do đó : A > 7/12 .

Vậy bài toán được chứng minh .

16 tháng 11 2018

1/

a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)

\(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)

\(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)

16 tháng 11 2018

b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993

2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993

2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993

2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993

2.(1 − 1/x+1) = 3984/1993

1 − 1/x + 1= 3984/1993 :2

1 − 1/x+1 = 1992/1993

1/x+1 = 1 − 1992/1993

1/x+1=1/1993

<=>x+1 = 1993

<=>x+1=1993

<=> x+1=1993

<=> x = 1993-1

<=> x = 1992

Bài 2: 

b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ\left(1\right)\)

\(\Leftrightarrow d\in\left\{1;-1\right\}\)

\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)

hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)

Bài 1: 

a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)

\(=75\cdot\left(-4\right)+603\)

\(=603-300=303\)

Bài 2: 

a) Vì tổng của hai số là 601 nên trong đó sẽ có 1 số chẵn, 1 số lẻ

mà số nguyên tố chẵn duy nhất là 2

nên số lẻ còn lại là 599(thỏa ĐK)

Vậy: Hai số nguyên tố cần tìm là 2 và 599

4 tháng 4 2021

b,Gọi ƯCLN(21n+4,14n+3)=d

21n+4⋮d ⇒42n+8⋮d

14n+3⋮d ⇒42n+9⋮d

(42n+9)-(42n+8)⋮d

1⋮d ⇒ƯCLN(21n+4,14n+3)=1

Vậy phân số 21n+4/14n+3 là phân số tối giản

 

\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)

\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)

\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)

Do đó: A=B

BT1: CMR: a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\) c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\) d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\) e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\) f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\) BT2: Tính tổng a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\) b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\) BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\) CMR: 1 < S <...
Đọc tiếp

BT1: CMR:

a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)

b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)

c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)

d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)

f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)

BT2: Tính tổng

a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)

BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)

CMR: 1 < S < 2

1
22 tháng 3 2017

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