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A=(2 + 2²+ 2³) + (2⁴ + 2⁵ +2⁶) +......+(2⁶¹+2⁶²+2⁶³)

A = 14 + 2³(2 + 2² + 2³) + .............+ 2⁶⁰(2 + 2² + 2³)

A=14+2³.14 + ..................+ 2⁶⁰ . 14

A= 14(2³+..... +2⁶⁰) chia hết cho 14 ( vì 14 chia hết cho 14)

Vậy A chia hết cho 14

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)

a) Ta có:

S = 1/5 + 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63

Ta thấy:

1/13 < 1/12 ; 1/14 < 1/12 ; 1/15 < 1/12

=> 1/13 + 1/14 + 1/15 < 1/12 + 1/12 + 1/12 = 1/12 . 3 = 1/4  (1)

1/61 < 1/60 ; 1/62 < 1/60 ; 1/63 < 1/60

=> 1/61 + 1/62 + 1/63 < 1/60 + 1/60 + 1/60 = 1/60. 3 = 1/20  (2)

 Từ (1) và (2)

=> 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63 < 1/4 + 1/20

=>S =  1/5 + 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63 < 1/4 + 1/20 + 1/5 = 5/20 + 1/20 + 4/20 = 10/20 = 1/2 (ĐPCM)

b) Ta có:

\(P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

\(2P-P=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^2}+...+\frac{1}{2^{19}}-\frac{1}{2^{19}}-\frac{1}{2^{20}}\)

\(P=1-\frac{1}{2^{20}}< 1\)

=> P < 1

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

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

\(A=\frac{1}{5}+\frac{1}{15}.3+\frac{1}{63}.3\)

\(A=\frac{1}{5}+\frac{1}{5}+\frac{1}{21}\)

\(A=\frac{47}{105}\)

Mà: \(\frac{47}{105}< \frac{47}{94}=\frac{1}{2}\)

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

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