a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)

\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)

\(=a^{12}\cdot b^3\)

b) \(b^6\cdot b\cdot c^7\cdot c^8\)

\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)

\(=b^7\cdot c^{15}\)

c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)

\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)

\(=a^{18}\cdot c^{21}\)

d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)

\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)

\(=a^5\cdot b^4\cdot c^4\)

a) Kiểm tra lại nhé

b) \(b^6.b^7.c^8\)

\(=b^{6+7}.c^8=b^{13}.c^8\)

c) \(a^8.a^9.a.c.c^{20}\)

\(=a^{8+9+1}.c^{1+20}\)

\(=a^{18}.c^{21}\)

d) \(a^2.a^3.b^4.c.c^3\)

\(=a^{2+3}.b^4.c^{1+3}\)

\(=a^5.b^4.c^4\)

\(#WendyDang\)

Câu b mình cũng không hiểu nha ( phần b mình làm bị sai á ).

\(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2B=2\cdot\left(2^{100}-2^{99}+2^{98}-...+2^2-2\right)\)

\(2B=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(2B+B=2^{101}-2^{100}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)

\(3B=2^{101}-2\)

\(B=\dfrac{2^{101}-2}{3}\)

\(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ =\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2\right)\\ =\left(2^2+...+2^{98}+2^{100}\right)-\left(2+...+9^{97}+9^{99}\right)\\ =M+N\left(1\right)\)

Xét \(M=2^2+...+2^{98}+2^{100}\\ 4M=2^4+...+2^{100}+2^{102}\\ 4M-M=2^4+...+2^{100}+2^{102}-2^2-...-2^{98}-2^{100}\\ 3M=2^{102}-2^2\\ M=\dfrac{2^{102}-2^2}{3}\left(2\right)\)

Xét \(N=2+...+2^{97}+2^{99}\\ 4N=2^3+...+2^{99}+2^{101}\\ 4N-N=2^3+...+2^{99}+2^{101}-2-...-2^{97}-2^{99}\\ 3N=2^{101}-2\\ N=\dfrac{2^{101}-2}{3}\left(3\right)\)

Từ `(1);(2)` và `(3)` suy ra 

\(B=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ =\dfrac{2^{102}-2^2-2^{101}+2}{3}=\dfrac{2^{101}\left(2-1\right)-2}{3}\\ =\dfrac{2^{101}-2}{3}\)

GIUP EMM DANG CAN GAPPP 

15 nhân 3 mũ 11 cộng 4 nhân 27 mũ 4 phần 9 mũ 7

\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{59}}.\)

\(E=2E-E=1-\frac{1}{2^{60}}\)

CÓ

\(3+3^2+3^3+...+3^{2012}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)

\(=40\left(3+...+3^{2009}\right)⋮40\)

​​

rrrrr

a) \(4.8^6.2.8^3\)

\(=2^2.\left(2^3\right)^6.2.\left(2^3\right)^3\)

\(=2^2.2^{18}.2.2^9\)

\(=2^{2+18+1+9}\)

\(=2^{30}\)

______

b) \(12^2.2.12^3.6\)

\(=12^2.12^3.2.6\)

\(=12^2.12^3.12\)

\(=12^{2+3+1}\)

\(=12^6\)

c) \(6^3.2.6^4.3\)

\(=6^3.6^4.2.3\)

\(=6^3.6^4.6\)

\(=6^{3+4+1}\)

\(6^8\)

a) \(4\cdot8^6\cdot2\cdot8^3\)

\(=2^2\cdot\left(2^3\right)^6\cdot2\cdot\left(2^3\right)^3\)

\(=2^2\cdot2^{18}\cdot2\cdot2^9\)

\(=2^{30}\)

b) \(12^2\cdot2\cdot12^3\cdot6\)

\(=12^2\cdot12\cdot12^3\)

\(=12^6\)

c) \(6^3\cdot2\cdot6^4\cdot3\)

\(=6^3\cdot6\cdot6^4\)

\(=6^8\)

Dòng cuối câu c phiền bạn thêm dấu " \(=\) "  vô trước số \(6^8\) nhé.

a) 5¹⁴ = 5^2.7= (5²)⁷= 25⁷

3²¹= 3^3.7= (3³)⁷= 27⁷

Mà 25 < 27 

=> 5¹⁴<3²¹

b) 4³³= (4³)¹¹  = 64¹¹

3⁴⁴= (3⁴)¹¹ = 81¹¹

=> 4³³<3⁴⁴

c) 2¹⁶¹ > 2¹⁶⁰ = (2⁴)⁴⁰= 16⁴⁰

Mà 13⁴⁰< 16⁴⁰ < 2¹⁶¹

=> 13⁴⁰<2¹⁶¹

d) 17¹⁵= 17^3.5= (17³)⁵ = 4913⁵

31¹⁰= (31²)⁵= 961⁵

Mà 4913 > 961

=> 17¹⁵> 31¹⁰

e) 2²⁵⁵ = (2⁵)⁴⁵ = 32⁴⁵

3¹⁸⁰ = (3⁴)⁴⁵ = 81⁴⁵

=> 2²⁵⁵ < 3¹⁸⁰

g) 202³⁰³ = (202³)¹⁰¹ = (101³.2³)¹⁰¹= (101³.8)¹⁰¹= (101².8.101)¹⁰¹= (101².808)¹⁰¹

303²⁰² = (303²)¹⁰¹= (3².101²)¹⁰¹= (101².9) ¹⁰¹

Vì 101².9 < 101².808

=> 202³⁰³>303²⁰²