3: \(C=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2014\cdot2016}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2015^2-1}\right)\)

\(=\dfrac{2^2-1+1}{2^2-1}\cdot\dfrac{3^2-1+1}{3^2-1}\cdot...\cdot\dfrac{2015^2-1+1}{2015^2-1}\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2015^2}{\left(2015-1\right)\left(2015+1\right)}\)

\(=\dfrac{2\cdot3\cdot...\cdot2015}{1\cdot2\cdot...\cdot2014}\cdot\dfrac{2\cdot3\cdot...\cdot2015}{3\cdot4\cdot...\cdot2016}\)

\(=\dfrac{2015}{1}\cdot\dfrac{2}{2016}=\dfrac{2015}{1008}\)

1: \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{50}\)

=>\(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 2-\dfrac{1}{50}\)

=>\(A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(2-\dfrac{1}{50}\right)\)

=>\(A< \dfrac{1}{2}-\dfrac{1}{200}\)

=>A<1/2

a: \(A=\dfrac{1}{2}:\dfrac{1}{20}-\dfrac{3}{5}\)

\(=\dfrac{1}{2}\cdot\dfrac{20}{1}-\dfrac{3}{5}\)

\(=10-\dfrac{3}{5}=\dfrac{47}{5}\)

b: \(B=1,2+2,5\cdot2-4,8\)

\(=1,2+5-4,8\)

=5-3,6=1,4

c: \(C=\dfrac{3}{13}\cdot\dfrac{9}{4}+\dfrac{3}{4}\cdot\dfrac{4}{-13}-\dfrac{3}{13}\cdot\dfrac{1}{4}+\dfrac{10}{13}\)

\(=\dfrac{3}{13}\left(\dfrac{9}{4}-\dfrac{4}{4}-\dfrac{1}{4}\right)+\dfrac{10}{13}\)

\(=\dfrac{3}{13}\cdot\dfrac{4}{4}+\dfrac{10}{13}=\dfrac{3}{13}+\dfrac{10}{13}=1\)

d: \(D=\left(3\dfrac{1}{3}+5\dfrac{2}{3}\right)\cdot\left(-2014^{12}\right)^0-30\%\cdot2,5\)

\(=\left(3+\dfrac{1}{3}+5+\dfrac{2}{3}\right)-0,3\cdot2,5\)

\(=9-0,75=8,25\)

Sửa đề: 7,15:0,5+7,15x9-0,715:0,1

=7,15x2+7,15x9-7,15

=7,15x(2+9-1)

=7,15x10=71,5

   7,15 : 0,5 + 7,15 x 9- 0,715:0,1

= 7,15 x 2 + 7,15 x 9 - 7,15

= 7,15 x ( 2 + 9 - 1 )

= 7,15 x 10

= 71,5

Nhận thấy:\(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},\dfrac{1}{4^2}< \dfrac{1}{3.4},...,\dfrac{1}{99^2}< \dfrac{1}{98.99},\dfrac{1}{100^2}< \dfrac{1}{99.100}\\ \)

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\\ \Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow A< 1-\dfrac{1}{100}\\ \Rightarrow A< \dfrac{99}{100}\)

Bạn xem lại xem có nhầm đề không nhỉ?

Diện tích hình thang ABCD là:

\(\dfrac{(4,4+11,2)\cdot5,8}{2}=45,24\mathit{(dm^2)}\)

Diện tích tam giác ADE là:

\(\dfrac12\cdot11,2\cdot6,4=35,84\mathit{(dm^2)}\)

Diện tích mảnh bìa là:

\(45,24+35,84=81,08\mathit{(dm^2)}\)

\(\rightarrow\) Chọn C. 81,08dm²

\(x-\dfrac{5}{3}.x=\dfrac{22}{9}\)

\(=>x.\left(1-\dfrac{5}{3}\right)=\dfrac{22}{9}\)

\(=>x.\left(\dfrac{3}{3}-\dfrac{5}{3}\right)=\dfrac{22}{9}\)

\(=>x.\left(-\dfrac{2}{3}\right)=\dfrac{22}{9}\)

\(=>x=\dfrac{22}{9}:\left(\dfrac{-2}{3}\right)=\dfrac{22}{9}\times\left(\dfrac{-3}{2}\right)\)

\(=>x=\dfrac{-11}{3}\)

\(#NqHahh\)

mk trả lời đc mỗi câu 3thui

Câu 3: số chẵn bé nhất có 4 c/s là 1000

           số chẵn lớn nhất có 4 c/s là 9998

 Có tất cả: (9998 - 1000) : 2 + 1 = 5000 ( số chẵn có 4 c/s )

                  Đ/s : 5000 số

s.o.s

ơ. bài này là của NTT đúng không

ko tính thời gian nghỉ,thời gian đi là :

75:50=1,5(H)=1h30'

Đến b vào số giờ là :

7H20'+20'+1h30'=9h10'

Thời gian di chuyển giữa 2 tỉnh A và B của xe ô tô là:

\(\dfrac{75}{50}-\dfrac{20}{60}=70\left(p\right)=1h10p\)

Ô tô đến B lúc :

\(7h20p+1h10p=8h30p\)

Đ/s:.....

\(\dfrac{1}{2}-\left(-\dfrac{2}{5}\right)+\dfrac{1}{3}+\dfrac{5}{7}-\left(-\dfrac{1}{6}\right)+\left(-\dfrac{4}{35}\right)+\dfrac{1}{41}\)

\(=\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{5}{7}+\dfrac{1}{6}+-\dfrac{4}{35}+\dfrac{1}{41}\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+-\dfrac{4}{35}+\dfrac{5}{7}\right)+\dfrac{1}{41}+\dfrac{1}{3}\)

\(=\dfrac{2}{3}+1+\dfrac{1}{3}+\dfrac{1}{41}\)

\(=\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+1+\dfrac{1}{41}\)

\(=1+1+\dfrac{1}{41}\)

\(=2+\dfrac{1}{41}\)

\(=\dfrac{83}{41}\)

\(#GP\)