Question

[Lê Phi Long]

 

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu 1:

1:

a: \(A=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)

b: \(1\cdot2+2\cdot3+...+2022\cdot2023\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+2022\left(1+2022\right)\)

\(=\left(1+2+...+2022\right)+\left(1^2+2^2+...+2022^2\right)\)

\(=\dfrac{2022\cdot2023}{2}+\dfrac{2022\left(2022+1\right)\left(2\cdot2022+1\right)}{6}\)

\(=1011\cdot2023+2022\cdot2023\cdot\dfrac{4045}{6}\)

\(=1011\cdot2023\left(1+2\cdot\dfrac{4045}{6}\right)\)

\(Q=\dfrac{1\cdot2+2\cdot3+...+2022\cdot2023}{2022\cdot2023\cdot2024}\)

\(=\dfrac{1011\cdot2023\left(1+\dfrac{4045}{3}\right)}{2022\cdot2023\cdot2024}=\dfrac{1}{2}\cdot\dfrac{1+\dfrac{4045}{3}}{2024}\)

\(=\dfrac{1}{2}\cdot\dfrac{4048}{3}:2024=\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{1}{3}\)

Câu 3:

1:

b: Số lần lấy được viên màu đỏ là 16 lần

=>Xác suất thực nghiệm là \(\dfrac{16}{40}=0,4\)

2: 2xy-4x-y=3

=>2x(y-2)-y+2=5

=>(2x-1)(y-2)=5

=>\(\left(2x-1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(1;7\right);\left(3;3\right);\left(0;-3\right);\left(-2;1\right)\right\}\)