\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

Ta có:

a) A = 2018 x 2020 = (2019 - 1) x (2019 + 1)

Áp dụng hằng đẳng thức thứ ba ta có:

A = 208 x 2020 = \(2019^2-1^2=2019^2-1\)

Vì \(2019^2-1< 2019^2\)

\(\Rightarrow\)A < B

b) A = \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1^2\right)\left(2^2+1^2\right)\left(2^4+1^2\right)\left(2^8+1^2\right)\left(2^{16}+1^2\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Vì \(2^{32}-1< 2^{32}\)

\(\Rightarrow\)A < B

a) Áp dụng hàng đăng thức (a - b) (a + b) = a² - b²

Ta có : A = 2018.2020 = (2019 - 1) (2019 + 1) = 2019² - 1

Mà B =  2019² 

Nên A < B 

Ta có : A = (2 + 1) (2² + 1) (2⁴ + 1) (2⁸ + 1)(2¹⁶ + 1) 

= (2² - 1) (2² + 1) (2⁴ + 1) (2⁸ + 1)(2¹⁶ + 1) 

= (2⁴ - 1) (2⁴ + 1) (2⁸ + 1)(2¹⁶ + 1) 

= (2⁸ - 1)(2⁸ + 1)(2¹⁶ + 1) 

= (2¹⁶ - 1) (2¹⁶ + 1) 

= 2³² - 1

Mà B = 2³²

Nên A < B 

\(A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{19.20}{2}}\)

=>    \(A=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)

=>   \(\frac{A}{2}=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

=>   \(\frac{A}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

=>    \(\frac{A}{2}=\frac{1}{3}-\frac{1}{20}\)

=>   \(\frac{A}{2}=\frac{20-3}{20.3}\)

=>   \(\frac{A}{2}=\frac{17}{60}\)

=>   \(A=\frac{17}{30}\)

VẬY    \(A=\frac{17}{30}\)

Ta có :\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+19}\)

\(=\frac{1}{3\times4}\times2+\frac{1}{4\times5}\times2+...+\frac{1}{19\times20}\times2\)

\(=2\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{19\times20}\right)=2\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\times\left(\frac{1}{3}-\frac{1}{20}\right)=2\times\frac{17}{60}=\frac{17}{30}\)

             A =\(\frac{1}{3x4:2}\)+ \(\frac{1}{4x5:2}\)+ ...+ \(\frac{1}{19x20:2}\)

             A = \(\frac{2}{3x4}\)+ \(\frac{2}{4x5}\)+...+ \(\frac{2}{19x20}\)

\(\frac{1}{2}\)x  A = \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+...+ \(\frac{1}{19x20}\)

  \(\frac{1}{2}\)x  A = \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+...+\(\frac{1}{19}\)- \(\frac{1}{20}\)

              A = ( \(\frac{1}{3}\)- \(\frac{1}{20}\)) : \(\frac{1}{2}\)

              A = \(\frac{17}{30}\)

Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho

mình bí

a , tổng các phân số đã cho là : 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 79/64 

b, \(\frac{79}{64}\)và \(\frac{2017}{2018}\)=  \(\frac{159422}{129152}\)và \(\frac{129088}{129152}\)= \(\frac{159422}{129152}\)> \(\frac{129088}{129152}\)

=> \(\frac{79}{64}\)> \(\frac{2017}{2018}\) 

a) 1/2 + 1/4 + 1/8 + 1/ 16 + 1/32 + 1/64 

=32/64 + 16/64 + 8/64 + 4/64 + 2/64

=32+16+8+4+2/64 = 66/64= 33/32

b) ta có 33/32 > 1 và 2017/2018<1

nên 33/32 > 2017/2018

Bạn khanh cuong ơi bài này tính đâu đơn giản thế đâu. Cậu làm vắn tắt vậy ai mà hiểu?