Là a chia cho hay a là đấy

la a chi cho do ban

Tổng các số trong trong ngoặc là:(100+1).[(100-1):1+1]:2=5050

  a=5050:5=1010

A= 51

B = 1

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

\(D=\left(-\dfrac{1}{3}\right)^1+\left(-\dfrac{1}{3}\right)^2+...+\left(-\dfrac{1}{3}\right)^{98}+\left(\dfrac{-1}{3}\right)^{99}\)

\(\Leftrightarrow\left(-\dfrac{1}{3}D\right)=\left(-\dfrac{1}{3}\right)^2+...+\left(-\dfrac{1}{3}\right)^{99}+\left(-\dfrac{1}{3}\right)^{100}\)

\(\Leftrightarrow D\cdot\dfrac{-4}{3}=\dfrac{1^{100}}{3^{100}}-\left(-\dfrac{1}{3}\right)=\dfrac{1}{3^{100}}+\dfrac{1}{3}=\dfrac{1+3^{99}}{3^{100}}\)

\(\Leftrightarrow D=\dfrac{3^{99}+1}{3^{100}}:\dfrac{-4}{3}=\dfrac{3^{99}+1}{-4\cdot3^{99}}\)

 A = ( 10 – 1).(100 – 2). (100 – 3) … (100 – n) với n = N* tích trên có đúng 100 thừa số

A = ( 10 – 1).(100 – 2). (100 – 3) … (100 – 100) = 99.98….0 = 0

\(A=\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)