A = -1/2 + 9999/9999 + (-2016)/2017 + (-8)/26 + (-10)/52

A = -1/2 + 1 + (-2016)/2017 + ( -8/26 + -10/52 )

A = -1/2 + 1 + (-2016)/2017 + (-1)/2

A = ( -1/2 + -1/2 ) +1 + (-2016)/2017

A = ( -1 ) + 1 + ( -2016)/2017

A = 0 + (-2016)/2017

A = -2016/2017

10 phan nha ban

Bạn giải chi tiết giúm mình đc k??

Câu 1.99²⁰và 9999¹⁰

⁼(99²)¹⁰=9801¹⁰

Vậy 9801¹⁰<9999¹⁰ suy ra  99²⁰<9999¹⁰

Câu 2. 3⁵⁰⁰và 7³⁰⁰

 3⁵⁰⁰=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=(7³)¹⁰⁰=343¹⁰⁰

Vậy 243¹⁰⁰<343¹⁰⁰ => 3⁵⁰⁰<7³⁰⁰

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

Ta có :

​Vì 101¹⁰>99¹⁰\(\Leftrightarrow\)101¹⁰.99¹⁰>99¹⁰.99¹⁰\(\Leftrightarrow\)9999¹⁰>99²⁰

Vậy 9999¹⁰>99²⁰

99^20 < 9999^10

Câu hỏi tương tự nha bạn

Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho

Bài 1 : 

a) 40/49 > 15/21

b) 22/49 > 3/8

c) 25/46 < 12/18

Bn uống nhầm thuốc hả Hà Chí Dương?

A= (1000+1)x1000:2=500500

B=1.2.3.4........11=39916800

=>A<B

tich ủng hộ nha