a) A = 4 + 4² + 4³ + 4⁴ + 4⁵ + 4⁶

A = ( 4 + 4² ) + ( 4³ + 4⁴ ) + ( 4⁵ + 4⁶ )

A = 4 . ( 1 + 4 ) + 4³ . ( 1 + 4 ) + 4⁵ . ( 1 + 4 )

A = 4 . 5 + 4³ . 5 + 4⁵ . 5

A = ( 4 + 4³ + 4⁵ ) . 5 \(⋮\)5 

b) tương tự

a: \(3\sqrt{200}=3\cdot10\sqrt{2}=30\sqrt{2}\)

b: \(-5\sqrt{50a^2b^2}=-5\cdot5\sqrt{2a^2b^2}\)

\(=-25\cdot\left|ab\right|\cdot\sqrt{5}\)

c: \(-\sqrt{75a^2b^3}\)

\(=-\sqrt{25a^2b^2\cdot3b}=-5\left|ab\right|\cdot\sqrt{3b}\)

Câu này đề hỏi gì vậy em?

a) \(12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)

\(=12\cdot\dfrac{4}{9}+\dfrac{4}{3}\)

\(=\dfrac{12\cdot4}{9}+\dfrac{4}{3}\)

\(=\dfrac{16}{3}+\dfrac{4}{3}\)

\(=\dfrac{16+4}{3}\)

\(=\dfrac{20}{3}\)

b) \(\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}\cdot\left(-\dfrac{1}{2}\right)^2\right]\)

\(=\dfrac{9}{4}-\left(\dfrac{1}{2}:2-9\cdot\dfrac{1}{4}\right)\)

\(=\dfrac{9}{4}-\left(\dfrac{1}{4}-9\cdot\dfrac{1}{4}\right)\)

\(=\dfrac{9}{4}-\dfrac{1}{4}\cdot\left(1-9\right)\)

\(=\dfrac{9}{4}+\dfrac{8}{4}\)

\(=\dfrac{17}{4}\) 

c) \(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\)

\(=-\dfrac{1}{12}:\dfrac{5}{11}+\dfrac{1}{12}\)

\(=\dfrac{1}{12}\cdot-\dfrac{11}{5}+\dfrac{1}{12}\)

\(=\dfrac{1}{12}\cdot\left(-\dfrac{11}{5}+1\right)\)

\(=\dfrac{1}{12}\cdot-\dfrac{6}{5}\)

\(=-\dfrac{1}{10}\) 

d) \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{4}{9}:\left(2+\dfrac{2}{3}\right)-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{9}{10}+\dfrac{1}{6}\)

\(=-\dfrac{11}{15}\) 

e) \(\dfrac{3^7\cdot8^6}{6^6\cdot\left(-2\right)^{12}}\)

\(=\dfrac{3^7\cdot\left(2^3\right)^6}{2^6\cdot3^6\cdot2^{12}}\)

\(=\dfrac{3^7\cdot2^{18}}{2^{6+12}\cdot3^6}\)

\(=\dfrac{2^{18}\cdot3^7}{2^{18}\cdot3^6}\)

\(=3^{7-6}\)

\(=3\)

\(a,12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\\ =12\cdot\dfrac{4}{9}+\dfrac{4}{3}\\ =\dfrac{16}{3}+\dfrac{4}{3}\\ =\dfrac{20}{3}\\ b,\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}.\left(-\dfrac{1}{2}\right)^2\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{2}\cdot\dfrac{1}{2}-9\cdot\dfrac{1}{4}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{8}{4}\right)\\ =\dfrac{17}{4}\)

\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =\left(-\dfrac{9}{12}+\dfrac{8}{12}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{3}{12}+\dfrac{4}{12}\right)\\ =-\dfrac{1}{12}\cdot\dfrac{11}{5}+\dfrac{1}{12}\\ =-\dfrac{11}{60}+\dfrac{1}{12}\\ =-\dfrac{1}{10}\)

\(d,\dfrac{-1^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left(-\dfrac{5}{6}\right)\\ =-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}+\dfrac{5}{6}\\ =-\dfrac{1}{15}+\dfrac{1}{6}+\dfrac{5}{6}\\ =\dfrac{1}{10}+\dfrac{5}{6}\\ =\dfrac{14}{15}\)

`e,` Không hiểu đề á c: )

Trả lời:

a, Ta có: 3²⁰ ; 27⁴ = ( 3³ )⁴ = 3¹²

Vì 3²⁰ > 3¹² nên 3²⁰ > 27⁴

b, 2²⁵ ; 16⁶ = ( 2⁴ )⁶ = 2²⁴

Vì 2²⁵ > 2²⁴ nên 2²⁵ > 16⁶ 

Trả lời:

c, 10³⁰ = ( 10³ )¹⁰  = 1000¹⁰ ; 4⁵⁰ = ( 4⁵ )¹⁰ = 1024¹⁰

Vì 1000¹⁰ < 1024¹⁰ nên 10³⁰ < 4⁵⁰

d, 5³⁴ ; 25.5³⁰ = 5² . 5³⁰ = 5³²

Vì 5³⁴ > 5³² nên 5³⁴ > 25.5³⁰

Trả lời:

e, 10³⁰ = ( 10³ )¹⁰ = 1000¹⁰ ; 2¹⁰⁰  = ( 2¹⁰ )¹⁰ = 1024¹⁰

Vì 1000¹⁰ < 1024¹⁰ nên 10³⁰ < 2¹⁰⁰

f, 5⁴⁰ = ( 5⁴ )¹⁰ = 625¹⁰ ; 620¹⁰ 

Vì 625¹⁰ > 620¹⁰ nên 5⁴⁰ > 620¹⁰ 

*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)

              \(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)

              \(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)

              \(=6\times\left(2^2+2^3+...+2^{2008}\right)\)

              \(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)

               \(\Rightarrow A⋮3\)

*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)

               \(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)

               \(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)

               \(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)

                \(\Rightarrow A⋮7\)

Mình sửa lại đề C 1 chút xíu

*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)

               \(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)

               \(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)

               \(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)

               \(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)

                \(\Rightarrow C⋮4\)

Các câu khác làm tương tự nhé. Chúc bạn học tốt!

Thanks bạn

Giải: 

A= 2 + 2 mũ 2 + 2 mũ 3 + 2 mũ 4 +....+ 2 mũ 2010

A= (2 + 2 mũ 2) + (2 mũ 3 + 2 mũ 4) +....+ (2 mũ 2009 + 2 mũ 2010)

A= 2(1 + 3) + 2 mũ 3 (1 + 2) + 2 mũ 2009 (1 +2_

A= 2.3 + 2 mũ 3.3 +....+ 2 mũ 2009.3

A= 3.(2 + 2 mũ 3 +....+ 2 mũ 2009) chia hết cho 3

A= (2 + 2 mũ 2 + 2 mũ 3) + (2 mũ 4 + 2 mũ 5 + 2 mũ 6) +....+ (2 mũ 2008 + 2 mũ 2009 + 2 mũ 2010)

A= 2(1 + 2 + 2 mũ 2) + 2 mũ 4(1+ 2 + 2 mũ 2) +...+ 2 mũ 2008.(1 + 2 + 2 mũ 2)

A= 2.7 + 2 mũ 4. 7 +.... + 2 mũ 2008.7

A= 7.(2 + 2 mũ 4 +....+ 2 mũ 22010 chia hết cho 7.

Các câu còn lại làm tương tự như câu a nha bạn!

2¹+2²+2³+......+2¹⁰

⇒2.(1+2)+2².(1+2)+......+2⁹.(1+2)

⇒2.3+2².3+........+2⁹.3

⇒3.(2+2²+....+2⁹)⋮3

Vậy 2¹+2²+2³+....+2¹⁰⋮3 (đpcm)

Đặt biểu thức trên là A, ta có:

\(A=2^1+2^2+2^3+...+2^{10}\)

\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(A=2^1.\left(1+2\right)+2^3.\left(1+2\right)+...+2^9.\left(1+2\right)\)

\(A=2^1.3+2^3.3+...+2^9.3\)

\(A=3.\left(2^1+2^3+...+2^9\right)\)

\(\Rightarrow A⋮3\)

2¹+2²+2³+....+2¹⁰

=2.(1+2)+2².(1+2)+........+2⁹.(1+2)

=2.3+2².3+...+2⁹.3

=3.(2+2²+....+2⁹)

vi 3⋮3=>3.(2+2²+....+2⁹)

vậy 2+2²+2³+...+2¹⁰ ⋮ 3