a: =>4y+15/16=1

=>4y=1/16

hay y=1/64

b: =>10y+1023/1024=1

=>10y=1/1024

hay  y=1/10240

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{19}\right)\left(1-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{18}{19}.\dfrac{19}{20}=\dfrac{1}{20}>\dfrac{1}{21}\)

2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)

\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)

\(B=\left(\dfrac{2^2}{2^2}-\dfrac{1}{2^2}\right)\cdot\left(\dfrac{3^2}{3^2}-\dfrac{1}{3^2}\right)....\left(\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\right)\)

\(B=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}....\cdot\dfrac{100^2-1}{100^2}\)

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

\(B=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}\)

\(B=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot101}{2^2\cdot3^2\cdot4^2\cdot5^2\cdot....\cdot100^2}\)

\(B=\dfrac{1\cdot101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)

\(B=\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)

Mà: \(\dfrac{1}{2}=\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\) 

Ta có: \(101< 3\cdot4\cdot5\cdot...\cdot100\)

\(\Rightarrow\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}< \dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)

\(\Rightarrow B< \dfrac{1}{2}\)     

a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)

\(=-\dfrac{1}{10}\)

9<10

=>1/9>1/10

=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)

=>\(A>-\dfrac{1}{9}\)

b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

20<21

=>\(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)

a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)

\(=-4\sqrt{5}+15\sqrt{2}\)

b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)

\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)

\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)

\(=8\sqrt{3}+2\sqrt{2}-4\)

c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)

\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)

=9-3

=6

d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

\(A=\left[\dfrac{1}{100}-1^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{3}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)\(=\left[\dfrac{1}{100}-1^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{3}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)Mà \(\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2=\dfrac{1}{100}-\dfrac{1}{100}=0\)

\(\Rightarrow A=0\)

\(\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)

\(=\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)

\(=\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...0...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)

\(=0\)

Vậy...

gọi a, b, c lần lượt là số cây trồng duoc cua cac lơp 7A, 7B, 7C. Ta co:
b = 8/9a => b/a = 8/9 => b/8 = a/9
=> b/16 = a/18 ....(1) (nhan ca hai ve cho 1/2)
c = 17/16b
=> c/17 = b/16 .....(2)
từ (1) và (2) ta được:
a/18 = b/16 = c/17 = (a+b+c)/(18+16+17) = 1020/51 = 20

=> a = 18.20 = 360 cây
b = 16.20 = 320 cây
c = 17.20= 340 cay

A = - 5²² - { - 222 - [ - 122 - (100 - 5²²) + 2022] }

A = - 5²² - { -222 - [- 122 - 100 + 5²² ] + 2022}

A = - 5²² - { -222 - { - 222 + 5²² } + 2022}

A = - 5²² - {- 222 + 222 - 5²² + 2022}

A = -5²² + 5²² - 2022

A = - 2022

B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)

B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2

B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2

B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)

B = \(\dfrac{2+3+4+...+21}{2}\)

B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)

B = \(\dfrac{23\times20:2}{2}\)

B = \(\dfrac{23\times10}{2}\)

B = 23 

Bài 2: 

a: \(=\sqrt{2}-\dfrac{2}{5}\sqrt{2}+2\sqrt{2}+2\sqrt{2}=\dfrac{23}{5}\sqrt{2}\)