\(\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot.....\cdot99}\\ =\dfrac{100}{2}=50\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{3^2}< \dfrac{1}{2\cdot3};.....;\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}\\ =>\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+....+\dfrac{1}{49\cdot50}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+1\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-.....-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{50}=2-0-0-0.....-\dfrac{1}{50}\\ =2-\dfrac{1}{50}< 2=>\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 2\)

\(\dfrac{13}{8}+\dfrac{9}{24}+\left(\dfrac{27}{8}+\dfrac{15}{24}\right)\\=\left(\dfrac{13}{8}+\dfrac{27}{8}\right)+\left(\dfrac{9}{24}+\dfrac{15}{24}\right)\\ =\dfrac{40}{8}+\dfrac{24}{24}\\ =5+1\\ =6\)

Chu vi cái sân hcn là: \(\left(\dfrac{2}{3}+38,4\right)\cdot2=\dfrac{1172}{15}\left(m\right)\)

Diện tích cái sân hcn là:  \(\dfrac{2}{3}\cdot38,4=25,6\left(m^2\right)\)

+) Để \(\dfrac{n-5}{n-3}\) có giá trị nguyên thì \(n-5⋮n-3\)

 \(n-5⋮n-3\\ =>n-5-\left(n-3\right)⋮n-3\\ =>n-5-n+3⋮n-3\\ =>-2⋮n-3\\ =>n-3\inƯ\left(-2\right)\\ Ư\left(-2\right)=\left\{-1;1;-2;2\right\}\\ TH1:n-3=-1=>n=2\\ TH2:n-3=1=>n=4\\ TH3:n-3=-2=>n=1\\ TH4:n-3=2=>n=5\)

\(M=5+5^2+5^3+5^4...+5^{80}\\ M=5\left(1+5\right)+5^3\left(1+5\right)+5^5\left(1+5\right)+.....+5^{79}\left(1+5\right)\\ M=5\cdot6+5^3\cdot6+5^5\cdot6+....+5^{79}\cdot6\\ M=6\left(5+5^3+5^5+5^7+....+5^{79}\right)⋮6=>M⋮6\)

\(3a=4b=>\dfrac{a}{4}=\dfrac{b}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nha, ta có

\(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{2a}{8}=\dfrac{3b}{9}=\dfrac{2a+3b}{8+9}=-\dfrac{36}{17}\)

Từ: \(\dfrac{a}{4}=-\dfrac{36}{17}=>a=-\dfrac{144}{17}\)

       \(\dfrac{b}{3}=-\dfrac{36}{17}=>b=-\dfrac{108}{17}\)

\(4x=3y=2z\\ =>\dfrac{4x }{12}=\dfrac{3y}{12}=\dfrac{2z}{12}\\ =>\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}\)

Áp dụng tích chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x+y+z}{3+4+6}=\dfrac{65}{13}=5\)

Từ: \(\dfrac{x}{3}=5=>x=15\)

     \(\dfrac{y}{4}=5=>y=20\)

\(\dfrac{z}{6}=5=>z=30\)

\(5^{27}=5^{3\cdot9}=125^9\\ 2^{63}=2^{7\cdot9}=512^7=128^9\\ 5^{28}=5^{7\cdot4}=625^7\)

Vì 125⁹ < 128⁹ => 5²⁷ < 2⁶³ 

Vì 512⁷<625⁷ => 2⁶³ < 5²⁸

=>  5²⁷ < 2⁶³ < 5²⁸

\(5⋮\left(x+1\right)\\ =>x+1\inƯ\left(5\right)\\ Ư\left(5\right)=\left\{-1;1;5;-5\right\}\\ TH1:x+1=-1\\ x=-2\\ TH2:x+1=1\\ x=0\\ TH3:x+1=5\\ x=4\\ TH4:x+1=-5\\ x=-6\\ =>x\in\left\{0;4;-6;-2\right\}\)