\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{11}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10}{11}\)

\(A=\frac{1\cdot2\cdot3\cdot...\cdot10}{2\cdot3\cdot4\cdot...\cdot11}=\frac{1}{11}\)

$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{{99}^2}\right)\left(1-\frac{1}{{100}^2}\right)$

$=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{99}\right)\left(1+\frac{1}{99}\right)\left(1-\frac{1}{100}\right)\left(1+\frac{1}{100}\right)$

$=\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}...\frac{98}{99}.\frac{100}{99}.\frac{99}{100}.\frac{101}{100}$

$=\frac{\mathrm{1.2...98.99}}{\mathrm{2.3...99.100}}.\frac{\mathrm{3.4...100.101}}{\mathrm{2.3...99.100}}$

$=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}$

chép mạng

Áp dụng công thức tính góc:\(\frac{n\left(n-1\right)}{2}\)= số góc

=>n(n-1)=1560.2

=>n.(n-1)=3120

(Vì n(n-1) là hai số nguyên liên tiếp nên ta chỉ cần tìm tích của 2 số nguyên liên tiếp là 3120)

=>n=.....

Tìm 2 số liên tiếp nhân với nhau bằng 3120 và n là số bé hơn nhé

25 - (30 + x) = x - (27 - 8)

25 - 30 - x = x - 27 + 8

- x - x = -25 + 30 - 27 + 8

-2x = -14

x = -14 : (-2)

x = 7.

\(\frac{2}{x}=\frac{y}{15}=\frac{\left(-25\right)}{75}\)

\(\Rightarrow\frac{y}{15}=\frac{\left(-1\right)}{3}\)

\(\frac{y}{15}=\frac{\left(-5\right)}{15}\)

\(\Rightarrow15y=15\times\left(-5\right)\)

\(15y=\left(-75\right)\)

\(y=\left(-75\right)\div15\)

\(y=\left(-5\right)\)

\(\Rightarrow\frac{2}{x}=\frac{\left(-5\right)}{15}\)

\(\frac{2}{x}=\frac{\left(-1\right)}{3}\)

\(\Rightarrow-x=3\times2\)

\(-x=6\)

\(x=\left(-6\right)\)

\(V\text{ậy}x=\left(-6\right);y=\left(-5\right)\)

|x + 2| - 13 = -1

=> |x + 2| = -1 + 13

=> |x + 2| = 12

=> \(\orbr{\begin{cases}x+2=12\\x+2=-12\end{cases}}\)

=> \(\orbr{\begin{cases}x=10\\x=-14\end{cases}}\)

x - (17 - x) = -7

=> x - 17  +x = -7

=> 2x - 17 = -7

=> 2x = -7 + 17

=> 2x = 10

=> x = 5

10(x + 7) - 8(x + 5) = 6(-5) + 24

=> 10x + 70 - 8x - 40 = -30 + 24

=> 2x + 30 = -6

=> 2x = -6 - 30

=> 2x = -36

=> x = -18

\(A=\frac{n+3}{n-2}=\frac{n-2+5}{n-2}=1+\frac{5}{n-2}\)

Để A nguyên khi \(\frac{5}{n+2}\inℤ\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Đây là phần làm

Vì 12 : 6 = 2 nên 5.2 = 10 . Vậy X = 10

\(\frac{A}{3}=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{91-91}{91.94}\)

\(\frac{A}{3}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}=1-\frac{1}{94}=\frac{93}{94}\Rightarrow A=\frac{3.93}{94}\)