2024 x 89 + 12 x 2024 - 2024

= 2024 x 89 + 12 x 2024 - 2024 x 1

= 2024 x (89 + 12 - 1)

= 2024 x 100

= 202400

2024 x 89 + 12 x 2024 - 2024

= 2024 x 89 + 12 x 2024 - 2024 x 1

= 2024 x (89 + 12 - 1)

= 2024 x 100

= 202400

giải dễ hiểu 2/2024+4/2024+6/2024+....+2022/2024

\(S=C^0_{2024}+\dfrac{1}{2}C^2_{2024}+\dfrac{1}{3}C^4_{2024}+\dfrac{1}{4}C^6_{2024}+...+\dfrac{1}{1013}C^{2024}_{2024}\)

Ta có :

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

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

\(=\dfrac{1}{n+1}C^{2k}_{n+1}\)

\(\Rightarrow S_n=\dfrac{1}{n+1}\Sigma^{2k}_{k=0}C^{2k}_{n+1}=\dfrac{1}{n+1}\left(\Sigma^{2k}_{k=0}C^{2k-1}_{n+1}-C^0_{n+1}\right)=\dfrac{2^{2n-1}-1}{n+1}\)

\(\Rightarrow S=\dfrac{2^{2025}-1}{1013}\)

S = C₀₂₀₂₄ + 12.C₂₀₂₄ + 13.C₂₀₂₄ + 14.C₂₀₂₄ + ... + 11013.C₂₀₂₄

= (C₀₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + (C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + ... + (C₂₀₂₄)

= 11014.C₂₀₂₄

= 11014.

jhvugb

a) 2024 . 33 + 2024 + 4048 . 33

= 2024 . 33 + 2024 . 1 + 2024 . 2 . 33

= 2024 . ( 33 + 1 + 66 )

= 2024 . 100

= 202400

2024×33+2024+4048×33=

\(\left(\right. 2024 + 4048 \left.\right) \times 33 + 2024\)

\(2024 + 4048 = 6072\)

\(6072 \times 33 = 200376\)

\(2024\): \(200376 + 2024 = 202400\)

b)
\(8 \times \frac{1}{6} + 8 \times \frac{2}{6} + 8 \times \frac{3}{6}\)
\(\frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{6}{6} = 1\)

\(8 \times 1 = 8\)

(y - 1)²⁰²⁴ + |\(x+y-1\)| = 0

Vì (y - 1)²⁰²⁴ ≥ 0 ∀ y; |\(x+y-1\)| ≥ 0 ∀ \(x;y\)

(y - 1)²⁰²⁴ + |\(x+y-1\)| = 0 khi và chỉ khi 

 y - 1 = 0 và \(x+y-1\) = 0

y - 1 = 0 Suy ra y = 1. thay y = 1 vào biểu thức \(x+y-1=0\) ta có:

\(x+1-1=0\) ⇒ \(x=0-1+1\) \(x=0\)

Vậy \(x=0;y=1\) thay vào biểu thức A= \(x^{2024}\) + y²⁰²⁴ ta được:

A = 0²⁰²⁴ + 1²⁰²⁴ = 0 + 1 = 1