1.

bạn ghi rất kho hiểu nên mình khuyên bạn ghi lại trên word rồi cop vào đây

1/ \(346-\left[124+30:\left(5^8:5^7\right).2^2.1018^0\right]\)

\(=346-\left[124+30:5.4.1\right]\)

\(=346-\left[124+6.4.1\right]\)

\(=346-\left[124+24\right]\)

\(=346-148\)

\(=198\)

2/ \(x^{17}=x\)

\(\Leftrightarrow x\in\left\{0;1\right\}\)

3/

a/ \(A=2+2^2+..........+2^{50}\)

\(\Leftrightarrow2A=2^2+2^3+..........+2^{51}\)

\(\Leftrightarrow2A-A=\left(2^2+2^3+.......+2^{51}\right)-\left(2+2^2+......+2^{51}\right)\)

\(\Leftrightarrow A=2^{51}-2\)

b,c tương tự

Câu 2:

\(2\cdot S=2+2^2+...+2^{2018}\)

=>\(S=2^{2018}-1\)

\(5\cdot2^{2017}>2\cdot2^{2017}=2^{2018}>2^{2018}-1\)

nên \(B< 5\cdot2^{2017}\)

Bài 1: Theo đề, ta có : a : 18 ( dư 12 ) ( a \(\in N\) )

\(\Rightarrow\) a : 2.9 ( dư 3+9 )

\(\Rightarrow\) a : 9 ( dư 3 )

Bài 2 : Theo đề, ta có : B = 6 + m + n + 12

B = ( m + n ) + ( 6 + 12 )

B = ( m + n ) + 18

Vì \(18⋮3\) nên khi ( m + n ) \(⋮\) 3 thì B \(⋮3\)

Ngược lại, khi ( m + n ) \(⋮̸\) 3 thì B \(⋮̸\) 3.

Bài 3:

Ta có : A = \(2+2^2+2^3+...+2^{49}+2^{50}\)

A = \(\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{49}+2^{50}\right)\)

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

A = \(2.3+2^3.3+...+2^{49}.3\)

A = \(3\left(2+2^3+...+2^{49}\right)\) \(⋮\) 3

Ta có : A = \(2+2^2+2^3+2^4+2^5+...+2^{49}+2^{50}\)

A = \(\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{46}+2^{47}+2^{48}+2^{49}+2^{50}\right)\)

A = \(2\left(1+2+2^2+2^3+2^4\right)+...+2^{46}\left(1+2+2^2+2^3+2^4\right)\)

A = 2 . 62 + ... + \(2^{46}.62\)

A = 62 ( 2 +...+ \(2^{46}\) )

A = 31 . 2( \(2+...+2^{46}\) ) \(⋮\) 31

Bài 4: Ta có : \(\overline{abcabc}\) = \(\overline{abc}000+\overline{abc}\) = \(\overline{abc}\left(1000+1\right)\) = \(\overline{abc}.1001\) = \(\overline{abc}.77.13\) \(⋮13\)

Vậy : \(\overline{abcabc}⋮13\)

Để mk làm bài 5 sau nha. Bây giờ đang bận

Bài 5:

a/ Ta có: \(n+5\) \(⋮\) n - 2 ( n \(\in\) N )

\(\Rightarrow\) n - 2 +7 \(⋮\) n - 2

\(\Rightarrow\) 7 \(⋮\) n - 2

\(\Rightarrow\) n - 2 \(\in\) Ư(7) = { 1 ; 7 }

\(\Rightarrow n\in\left\{3;9\right\}\)

b/ Ta có : 2n + 7 \(⋮\) n + 1 ( n \(\in\) N )

\(\Rightarrow\) 2( n + 1 ) + 5 \(⋮\) n + 1

\(\Rightarrow\) 5 \(⋮\) n + 1

\(\Rightarrow\) n + 1 \(\in\) Ư (5) = { 1 ; 5 }

\(\Rightarrow\) n \(\in\) { 0 ; 4 }

Chúc bn hc tốt!!!

a) 16¹⁹ và 8²⁵ 

Ta có :

16¹⁹ = ( 2⁴ )¹⁹ = 2⁷⁶

8²⁵ = ( 2³ )²⁵ = 2⁷⁵

Vì 2⁷⁶ > 2⁷⁵ Nên 16¹⁹ > 8²⁵

b) 5³⁶ và 11²⁴

Ta có :

5³⁶ = ( 5³ )¹² = 125¹²

11²⁴ = ( 11² )¹² = 121¹²

Vì 125¹² > 121¹² Nên 5³⁶ > 11²⁴

1.

\(M=3^0+3^1+......+3^{50}.\)

\(\Rightarrow3M=3+3^2+.......+3^{51}\)

\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)

\(\Rightarrow2M=3^{51}-1\)

\(\Rightarrow M=\frac{3^{51}-1}{2}\)

2.

\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)

                     \(8^{25}=\left(2^3\right)^5=2^{75}\)

Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)

\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

                      \(11^{24}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

1)Ta thấy: \(\dfrac{1}{n^2}=\dfrac{1}{n.n}< \dfrac{1}{\left(n-1\right)n}\)

=>A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49.50}\)

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

Vậy A<2

2)Ta có:2S=6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)

2S-S=(6+3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\))-(3+\(\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\))

=>S=6-\(\dfrac{3}{2^9}=\dfrac{6.2^9-3}{2^9}\)

Vậy S=\(\dfrac{6.2^9-3}{2^9}\)

Các bạn cố giúp mink nhé mai mình phải nộp rồi

Lời giải:

Câu 1:

Dễ thấy:

\(\dfrac{1}{1^2}=1\)

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(....\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

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

\(\Rightarrow A< 1+1-\dfrac{1}{50}\)

\(\Rightarrow A< 2-\dfrac{1}{50}\)

\(\Rightarrow A< 2\)

Câu 2:

Ta có: \(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+\dfrac{3}{2^3}+...+\dfrac{3}{2^9}\)

\(\Rightarrow2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)

\(\Rightarrow2S-S=S=6-\dfrac{3}{2^9}=6-\dfrac{3}{512}=\dfrac{3069}{512}\)

\(\Rightarrow S=\dfrac{3069}{512}\)