\(C=2018^{2011}+\dfrac{1}{2018^{2019}+1}\)

\(D=\dfrac{2018^{2017}}{2018^{2013}+1}=\dfrac{2018^{2013}.2018^4}{2018^{2013}+1}=\dfrac{\left(2018^{2013}+1-1\right).2018^4}{2018^{2013}+1}=2018^4-\dfrac{2018^4}{2018^{2013}+1}\)

mà \(2018^4< 2018^{2011}\)

\(\Rightarrow D=2018^4-\dfrac{2018^4}{2018^{2013}+1}< 2018^{2011}-\dfrac{2018^4}{2018^{2013}+1}\)

mà \(2018^{2011}-\dfrac{2018^4}{2018^{2013}+1}< C=2018^{2011}+\dfrac{1}{2018^{2019}+1}\)

\(\Rightarrow D< C\)

\(A=\dfrac{3n+2}{n+1}=\dfrac{3n+3-2}{n+1}=\dfrac{3\left(n+1\right)-2}{n+1}=3-\dfrac{2}{n+1}\)

Để A có giá trị nguyên ⇒ n+1 là Ư(2)={-1;1;-2;2}

⇒ n+1 ϵ {-1;1;-2;2}

⇒ n ϵ {-2;0;-3;1}

Diện tích hình thang :

\(\dfrac{1}{2}x\left(\dfrac{4}{9}+\dfrac{3}{7}\right)x\dfrac{2}{5}=\dfrac{1}{2}x\dfrac{35}{63}x\dfrac{2}{5}=\dfrac{1}{9}\) (m²)

a) \(-2017\le x\le2018\)

\(\Rightarrow x\in\left\{-2017;-2016;...0;1;2;...2018\right\}\)

\(\Rightarrow\left(-2017\right)+\left(-2016\right)+...+0+1+2;...+2018=2018\)

b) \(a+3\le x\le a+2018\) \(\left(a\in N\right)\)

\(\Rightarrow x\in\left\{a+3;a+4;...a+2018\right\}\)

\(\Rightarrow T=a+3+a+4+...+a+2018\)

\(\Rightarrow T=a+a+...+a+3+4+...2018\)

\(\Rightarrow T=2016a+2016=2016\left(a+1\right)\)

\(A=1-2+3-4+5-6+7-8+...+99-100\)

\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

\(A=\left(-1\right).50\)

\(A=-50\)

\(B=1+3-5-7+9+11-...-397-399\)

\(B=1-2+2-2+2-...+2-2-399\)

\(B=1-399\)

\(B=-398\)

\(C=1-2-3+4+5-6-7+...+97-98-99+100\)

\(C=-1+1-1+1-...-1+1\)

\(C=0\)

\(D=2^{2024}-2^{2023}-...-1\)

\(D=2^{2024}-\left(2^0+2^1+2^2+...2^{2023}\right)\)

\(D=2^{2024}-\left(\dfrac{2^{2024}-1}{2-1}\right)\)

\(D=2^{2024}-\left(2^{2024}-1\right)\)

\(D=2^{2024}-2^{2024}+1\)

\(D=1\)

A) Từ 1 đến 9 : (9+1):2=5 (chữ số)

    Từ 10 đến 99 : (89+1):2=45 (chữ số)

    Từ 100 đến 999 : (899+1):2=450 (chữ số)

    Từ 1000 đến 1999 : (999+1):2=500 (chữ số)

Tổng chữ số : 5+45+450+500= 1000 (chữ số).

B) Vì từ 1 đến 1999 có 1999 số nên không có chữ số thứ 2000 của dãy.

\(A=12\dfrac{2}{5}.\left(\dfrac{-7}{3}\right)-3\dfrac{2}{5}.\left(\dfrac{-7}{3}\right)\)

\(A=\dfrac{62}{5}.\left(\dfrac{-7}{3}\right)-\dfrac{17}{5}.\left(\dfrac{-7}{3}\right)\)

\(A=\left(\dfrac{-7}{3}\right).\left(\dfrac{62}{5}-\dfrac{17}{5}\right)\)

\(A=\left(\dfrac{-7}{3}\right).\dfrac{45}{5}\)

\(A=-21\)

\(B=\left(\dfrac{2}{3}\right)^3:\left(\dfrac{2}{3}\right)^2+\left(-1\dfrac{1}{2}\right):150\%\)

\(B=\left(\dfrac{2}{3}\right)^1-\dfrac{3}{2}:1,5\)

\(B=\dfrac{2}{3}-\dfrac{3}{2}:\dfrac{3}{2}\)

\(B=\dfrac{2}{3}-1\)

\(B=-\dfrac{1}{3}\)

Số phần phân số còn lại là :

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

Số sách trong thùng ban đầu là :

\(12:\dfrac{1}{2}=12.\dfrac{2}{1}=24\) (quyển sách)

Câu 1 : 

\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)

Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)

mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)

\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)

Câu 2 :

\(\dfrac{2}{3}\&\dfrac{5}{7}\)

\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)

Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)