chờ tí

6 x 2022/9 + 2022/9 x 4 - 2022/9

=6 x 2022/9 + 2022/9 x 4 - 2022/9x1

=2022/9x(6+4-1)

=2022/9x9=2022

c

=522 :5021 2022 :521
Có viết thiếu j ko ạ, mk ko hiểu lắm

125-12(x-8)=522:5021 2022:521

125-12*(x-8)=0.10396335391 3.88099808061

125-12*(x-8)=0.40348157697

12*(x-8)=125+0.40348157697

12*(x-8)=125.403481577

x-8=125.403481577:12

x-8=10.4502901314

x=10.4502901314+8

x=18.4502901314

\(2022-\left(\dfrac{1}{4}\right)^2\cdot4^2=2022-\left(\dfrac{1}{4}\cdot4\right)^2=2022-1^2=2021\)

2022 - (1/4)^2x=4^2

           (1/4)^2x=2022 - 4^2

           (1/4)^2x=2006

                      x=2006/(1/4)

                      x=8024

\(\dfrac{C_n^k}{\left(k+1\right)\left(k+2\right)}=\dfrac{n!}{\left(k+1\right)\left(k+2\right).k!\left(n-k\right)!}=\dfrac{1}{\left(n+1\right)\left(n+2\right)}.\dfrac{\left(n+2\right)!}{\left(n+2-\left(k+2\right)\right)!\left(k+2\right)!}\)

\(=\dfrac{1}{\left(n+1\right)\left(n+2\right)}.C_{n+2}^{k+2}\)

Đặt tổng trên là A

\(\Rightarrow A=\dfrac{-1.C_{2024}^3}{2023.2024}+\dfrac{2.C_{2024}^4}{2023.2024}+\dfrac{-3.C_{2024}^5}{2023.2024}+...+\dfrac{2022.C_{2024}^{2024}}{2023.2024}\)

\(=\dfrac{1}{2023.2024}\left(-1.C_{2024}^3+2.C_{2024}^4+...+2022.C_{2024}^{2024}\right)=\dfrac{1}{2023.2024}.B\)

Xét \(C=-2.\left(-C_{2024}^3+C_{2024}^4-C_{2024}^5+...+C_{2024}^{2024}\right)\)

\(\Rightarrow B-C=-3C_{2024}^3+4C_{2024}^4-5C_{2024}^5+...+2024.C_{2024}^{2024}\)

Ta có:

\(k.C_n^k=\dfrac{n!.k}{\left(n-k\right)!.k!}=n.\dfrac{\left(n-1\right)!}{\left(\left(n-1\right)-\left(k-1\right)\right)!.\left(k-1\right)!}=n.C_{n-1}^{k-1}\)

\(\Rightarrow B-C=-2024.C_{2023}^2+2024C_{2023}^3+...+2024.C_{2023}^{2023}\)

\(=-2024\left(C_{2023}^2-C_{2023}^3+...-C_{2023}^{2023}\right)\)

Xét khai triển:

\(\left(1-x\right)^k=C_k^0-xC_k^1+x^2C_k^2+...+\left(-1\right)^kx^k.C_k^k\)

Thay \(k=2024\); \(x=1\)

\(\Rightarrow0=C_{2024}^0-C_{2024}^1+C_{2024}^2-C_{2024}^3+...+C_{2024}^{2024}\)

\(\Rightarrow-C_{2024}^3+...+C_{2024}^{2024}=C_{2024}^1-C_{2024}^2-1\)

\(\Rightarrow C=-2\left(C_{2024}^1-C_{2024}^2-1\right)=-2\left(2023-C_{2024}^2\right)\)

Thay \(k=2023;x=1\)

\(\Rightarrow0=C_{2023}^0-C_{2023}^1+C_{2023}^2+...-C_{2023}^{2023}\)

\(\Rightarrow C_{2023}^2-C_{2023}^3+...-C_{2023}^{2023}=C_{2023}^1-1=2022\)

\(\Rightarrow B-C=-2024.2022\)

\(\Rightarrow B=C-2022.2024=-2\left(2023-C_{2024}^2\right)-2022.2024\)

\(=-2.2023+2023.2024-2022.2024\)

\(=-2022\)

\(\Rightarrow A=\dfrac{-2022}{2023.2024}\)

A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022

A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)

A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)

A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]

A=674.(−1013,5)D=674.(-1013,5)

A=−683099

A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022

A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)

A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)

A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]

A=674.(−1013,5)D=674.(-1013,5)

A=−683099

2:

b=2000*2004

=(2002-2)*(2002+2)

=2002^2-4

=>b<a

1:

a: \(=8\cdot9\left(14+17+19\right)=72\cdot50=3600\)

Bài 1:

\(8\times9\times14+6\times17\times12+19\times4\times18\)

\(=8\times9\times14+3\times2\times17\times2\times2\times3+19\times4\times2\times9\)

\(=8\times9\times14+17\times8\times9+19\times8\times9\)

\(=8\times9\times\left(14+17+19\right)\)

\(=8\times9\times50\)

\(=72\times5\times10\)

\(=360\times10\)

\(=3600\)

Bài 2:

Ta có:

\(a=2022\times2022\)

Và: \(b=2000\times2004\)

Mà: \(2022>2000,2022>2004\)

\(\Rightarrow2022\times2022>2000\times2004\)

\(\Rightarrow a>b\)

\(\dfrac{-3}{7}\)(\(\dfrac{5}{9}\)+\(\dfrac{4}{9}\)) + 0

=\(\dfrac{-3}{7}\)

*Lần sau bạn viết rõ đề ra, để thế này nhiều ng sẽ không hiểu!

\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(2022\right)^0\)

\(=\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+1\)

\(=\dfrac{-3}{7}.1+1\)

\(=\dfrac{-3}{7}+1\)

\(=\dfrac{-3+7}{7}\)

\(=\dfrac{4}{7}\)

Đáp số là 2021/2022 em nhé

\(\dfrac{2021}{2022}.\dfrac{7}{16}+\dfrac{9}{16}.\dfrac{2021}{2022}=\dfrac{2021}{2022}\left(\dfrac{7}{16}+\dfrac{9}{16}\right)=\dfrac{2021}{2022}.1=\dfrac{2021}{2022}\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu để của bạn hơn nhé.