Question

Tính hợp lí

a) \(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

b) 1 + 2 + 2² + 2³ + 2⁴ +...+ 2⁹⁹ + 2¹⁰⁰

giúp mình với huuhu

Answer 1

a) \(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

= \(10.\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

= \(10.\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

= \(10.\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}.\left(2.3-1\right)}\)

= \(10.\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)

= \(10.\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)

= \(10.\dfrac{2^{12}.3^{10}.2.3}{2^{11}.3^{11}.5}\)

=\(10.\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)

= \(10.\dfrac{2^2.1}{1.1.5}\)

= \(10.\dfrac{4}{5}\)

= \(2.\dfrac{4}{1}\)

= 8

Answer 2

b) 1 + 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

Đặt S = 1 + 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

2S = 2 ( 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰ )

= 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹

2S - S = ( 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹ ) - ( 1 + 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ )

= 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ + 2¹⁰¹ - 1 - 2 - 2² - 2³ - 2⁴ - ... - 2¹⁰⁰

= 2¹⁰¹ - 1 = S

Vậy 1 + 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰ = 2¹⁰¹ - 1

Answer 3

b) Đặt A vào biểu thức, ta có :

A= 1+ 2 + 2² + 2³ + 2⁴ +...+ 2⁹⁹ + 2¹⁰⁰

2A= 2 + 2² +2³ + 2⁴ + 2⁵ +...+ 2¹⁰⁰ + 2²⁰¹

2A - A= 2²⁰¹ - 1

Answer 4

CÂU B đặt biểu thức cần tìm là A

ta tính 2A

rồi lấy 2A trừ đi A =2^101-1