\(A=\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\) 

\(A=\dfrac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{11}.3+\left(2.3\right)^{11}}\) 

\(A=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{11}.3+2^{11}.3^{11}}\) 

\(A=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}.2.3+2^{11}.3^{11}}\) 

\(A=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}.6+2^{11}.3^{11}}\)  

\(A=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(6+1\right)}\)  

\(A=\dfrac{2.6}{3.7}\)

\(A=\dfrac{12}{21}=\dfrac{4}{7}\)

Giải:

Ta có:

\(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\) 

\(B=1+\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)\) 

\(B=\dfrac{2021}{2021}+\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}\) 

\(B=2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{\left[2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\right]}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=2021\) 

Vậy \(\dfrac{A}{B}=2021\)

\(\left(1\dfrac{2}{3}\right).\left(1\dfrac{2}{5}\right).\left(1\dfrac{2}{7}\right).....\left(1\dfrac{2}{2011}\right).\left(1\dfrac{2}{2013}\right)\) 

\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}.....\dfrac{2013}{2011}.\dfrac{2015}{2013}\) 

\(=\dfrac{5.7.9.....2013.2015}{3.5.7.....2011.2013}\) 

\(=\dfrac{2015}{3}\)

\(x-\left(\dfrac{4}{6}-2^3\right)=4.6\) 

       \(x-\dfrac{2}{3}+8=24\) 

                     \(x=24-8+\dfrac{2}{3}\) 

                     \(x=\dfrac{50}{3}\) 

Giải:

\(\dfrac{n+7}{n+8}\) 

Gọi \(ƯCLN\left(n+7;n+8\right)=d\) 

\(\Rightarrow\left[{}\begin{matrix}n+7⋮d\\n+8⋮d\end{matrix}\right.\)   

\(\Rightarrow\left(n+8\right)-\left(n+7\right)⋮d\) 

\(\Rightarrow1⋮d\) 

\(\Rightarrow d=1\) 

Vậy \(\dfrac{n+7}{n+8}\) là p/s tối giản

\(\dfrac{4n+7}{n+2}\) 

Gọi \(ƯCLN\left(4n+7;n+2\right)=d\) 

\(\Rightarrow\left[{}\begin{matrix}4n+7⋮d\\n+2⋮d\end{matrix}\right.\)   \(\Rightarrow\left[{}\begin{matrix}4n+7⋮d\\4.\left(n+2\right)⋮d\end{matrix}\right.\)   \(\Rightarrow\left[{}\begin{matrix}4n+7⋮d\\4n+8⋮d\end{matrix}\right.\)

\(\Rightarrow\left(4n+8\right)-\left(4n+7\right)⋮d\)

\(\Rightarrow1⋮d\) 

\(\Rightarrow d=1\) 

Vậy \(\dfrac{4n+7}{n+2}\) là p/s tối giản

\(\dfrac{5n+12}{3n+7}\) 

Gọi \(ƯCLN\left(5n+12;3n+7\right)=d\) 

\(\Rightarrow\left[{}\begin{matrix}5n+12⋮d\\3n+7⋮d\end{matrix}\right.\)   \(\Rightarrow\left[{}\begin{matrix}3.\left(5n+12\right)⋮d\\5.\left(3n+7\right)⋮d\end{matrix}\right.\)   \(\Rightarrow\left[{}\begin{matrix}15n+36⋮d\\15n+35⋮d\end{matrix}\right.\) 

\(\Rightarrow\left(15n+36\right)-\left(15n+35\right)⋮d\) 

\(\Rightarrow1⋮d\) 

\(\Rightarrow d=1\) 

Vậy \(\dfrac{5n+12}{3n+7}\) là p/s tối giản

Chúc bạn học tốt!

A={x∈N*|x≤900}

Chọn khẳng định đúng trong các khẳng định dưới đây:

Hình tạo bởi hai tia bất kì trên một đường thẳng là góc bẹt

Hình tạo bởi hai tia có nhiều điểm chung (nhưng không trùng nhau) là một góc bẹt.

Hình tạo bởi hai tia trùng nhau là một góc bẹt

Hình tạo bởi tia cắt nhau là góc bẹt.

Cho n + 1 tia chung gốc. Hỏi tạo được \(\dfrac{\left(n+1\right).n}{2}\) góc?

\(5^{13}:5^{10}-25.2^2=5^3-5^2.4=5^2.5-5^2.4=25.\left(5-4\right)=25.1=25\) 

A={x∈N*|x<901}