5) a) Sửa: \(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\right]\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\cdot2\right]\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\cdot2\right]\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\right]\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right)\right]\)

\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right)\)

\(A=1\)

5) b) \(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\) 

\(B=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{9}{10}\cdot...\cdot\dfrac{779}{780}\)

\(B=\dfrac{2\cdot2}{3\cdot2}\cdot\dfrac{5\cdot2}{6\cdot2}\cdot\dfrac{9\cdot2}{10\cdot2}\cdot...\cdot\dfrac{779\cdot2}{780\cdot2}\)

\(B=\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\cdot...\cdot\dfrac{1558}{1560}\)

\(B=\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot\dfrac{3\cdot6}{4\cdot5}\cdot...\cdot\dfrac{38\cdot41}{39\cdot40}\)

\(B=\dfrac{1\cdot4\cdot2\cdot5\cdot3\cdot6\cdot...\cdot39\cdot41}{2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot39\cdot40}\)

\(B=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot38\right)\cdot\left(4\cdot5\cdot6\cdot...\cdot41\right)}{\left(2\cdot3\cdot4\cdot...\cdot39\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot40\right)}\)

\(B=\dfrac{1\cdot41}{39\cdot3}\)

\(B=\dfrac{41}{117}\)

mình cần gấp lắm ạ pls

Gọi d là ƯCLN(21n+4, 14n+3) khi đó:

21n + 4 ⋮ d và 14n + 3 ⋮ d

⇒ 2(21n+4) ⋮ d và 3(14n + 3) ⋮ d

⇒ 42n + 8 ⋮ d và 42n + 9 ⋮ d

⇒ (42n + 9) - (42n + 8) ⋮ d

⇒ 42n + 9 - 42n - 8 ⋮ d

⇒ 1 ⋮ d

⇒ d = 1

Hay: ƯCLN(21n+4, 14n+3) = 1

Vậy phân số `(21n+4)/(14n+3)` là phân số tối giản

\(2=2;3=3;5=5\)

=>\(BCNN\left(2;3;5\right)=2\cdot3\cdot5=30\)

=>Các số có 4 chữ số không là bội của 60 nhưng chia hết cho 2;3;5 là 1050;1110;...;9990

Số số có 4 chữ số thỏa mãn yêu cầu đề bài là:

\(\dfrac{9990-1050}{60}+1=150\left(số\right)\)

bài này ở đâu vậy/?

hơi dài

-1552

Bài 2:

a) \(\dfrac{5}{6}:x=\dfrac{13}{9}\cdot\dfrac{3}{26}\)

\(\Rightarrow\dfrac{5}{6}:x=\dfrac{3}{9\cdot2}\)

\(\Rightarrow\dfrac{5}{6}:x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{5}{6}:\dfrac{1}{6}\)

\(\Rightarrow x=5\) 

b) \(\dfrac{x-1}{21}=\dfrac{3}{x+1}\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)=3\cdot21\)

\(\Rightarrow x^2-x+x-1=63\)

\(\Rightarrow x^2-1=63\)

\(\Rightarrow x^2-64=0\)

\(\Rightarrow x^2-8x+8x-64=0\)

\(\Rightarrow x\left(x-8\right)+8\left(x-8\right)=0\)

\(\Rightarrow\left(x+8\right)\left(x-8\right)=0\)

TH1: `x+8=0=>x=-8`

TH2: `x-8=0=>x=8` 

c) \(2\dfrac{1}{2}x+x=2\dfrac{1}{17}\)

\(\Rightarrow\dfrac{5}{2}x+x=\dfrac{35}{17}\)

\(\Rightarrow x\cdot\left(\dfrac{5}{2}+1\right)=\dfrac{35}{17}\)

\(\Rightarrow x\cdot\dfrac{7}{2}=\dfrac{35}{17}\)

\(\Rightarrow x=\dfrac{35}{17}:\dfrac{7}{2}\)

\(\Rightarrow x=\dfrac{10}{17}\)
d) \(x-3\dfrac{1}{2}x=-2\dfrac{6}{7}\)

\(\Rightarrow x-\dfrac{7}{2}x=-\dfrac{20}{7}\)

\(\Rightarrow x\cdot\left(1-\dfrac{7}{2}\right)=\dfrac{-20}{7}\)

\(\Rightarrow x\cdot\dfrac{-5}{2}=\dfrac{-20}{7}\)

\(\Rightarrow x=\dfrac{-20}{7}:\dfrac{-5}{2}\)

\(\Rightarrow x=\dfrac{8}{7}\) 

Bài 2:

a: \(\dfrac{5}{6}:x=\dfrac{13}{9}\cdot\dfrac{3}{26}\)

=>\(\dfrac{5}{6}:x=\dfrac{13}{26}\cdot\dfrac{3}{9}=\dfrac{1}{6}\)

=>\(x=\dfrac{5}{6}:\dfrac{1}{6}=\dfrac{5}{6}\cdot\dfrac{6}{1}=5\)

b: 

ĐKXĐ: x<>-1

\(\dfrac{x-1}{21}=\dfrac{3}{x+1}\)

=>\(\left(x-1\right)\left(x+1\right)=3\cdot21\)

=>\(x^2-1=63\)

=>\(x^2=64\)

=>\(\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

c: \(2\dfrac{1}{2}\cdot x+x=2\dfrac{1}{17}\)

=>\(x\left(2,5+1\right)=\dfrac{35}{17}\)

=>\(x\cdot\dfrac{7}{2}=\dfrac{35}{17}\)

=>\(x=\dfrac{35}{17}:\dfrac{7}{2}=\dfrac{35}{17}\cdot\dfrac{2}{7}=\dfrac{10}{17}\)

d: \(x-3\dfrac{1}{2}x=-2\dfrac{6}{7}\)

=>\(x\left(1-3,5\right)=-\dfrac{20}{7}\)

=>\(x\cdot2,5=\dfrac{20}{7}\)

=>\(x=\dfrac{8}{7}\)

\(5^{3x-2}+3\dfrac{1}{2}\cdot5^5=28,5\cdot5^5\)

=>\(5^{3x-2}=28,5\cdot5^5-3,5\cdot5^5\)

=>\(5^{3x-2}=5^5\cdot25=5^7\)

=>3x-2=7

=>3x=9

=>x=3

\(A=\dfrac{7\cdot9+14\cdot27+21\cdot36+28\cdot45}{21\cdot27+42\cdot81+63\cdot108+84\cdot135}\)

\(=\dfrac{7\cdot9+14\cdot27+21\cdot36+28\cdot45}{9\cdot7\cdot9+9\cdot14\cdot27+9\cdot21\cdot36+9\cdot28\cdot45}\)

\(=\dfrac{7\cdot9+14\cdot27+21\cdot36+28\cdot45}{9\cdot\left(7\cdot9+14\cdot27+21\cdot36+28\cdot45\right)}=\dfrac{1}{9}\)

\(B=2023^0-\dfrac{7}{9}=1-\dfrac{7}{9}=\dfrac{2}{9}\)

=>A<B