\(\dfrac{7}{13}.\dfrac{5}{19}+\dfrac{7}{19}.\dfrac{8}{13}-3.\dfrac{7}{19}\)

\(=\dfrac{7}{13.19}.\left(5+8\right)-3.\dfrac{7}{19}\)

\(=\dfrac{7}{13.19}.13-\dfrac{21}{19}\)

\(=\dfrac{7}{19}-\dfrac{21}{19}\)

\(=-\dfrac{14}{19}\)

\(\dfrac{7}{13}\cdot\dfrac{5}{19}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\)

\(=\dfrac{7}{19}\cdot\dfrac{5}{13}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\)

\(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-3\right)\)

\(=\dfrac{7}{19}\cdot\left(-2\right)=-\dfrac{14}{19}\)

Yes

Đúng nhé bạn!

\(\dfrac{A}{2024}=\dfrac{2024^{30}+1}{2024^{30}+2024}=1-\dfrac{2023}{2024^{30}+2024}\)

\(\dfrac{B}{2024}=\dfrac{2024^{29}+1}{2024^{29}+2024}=1-\dfrac{2023}{2024^{29}+2024}\)

\(2024^{30}+2024>2024^{29}+2024\)

=>\(\dfrac{2023}{2024^{30}+2024}< \dfrac{2023}{2024^{29}+2024}\)

=>\(-\dfrac{2023}{2024^{30}+2024}>-\dfrac{2023}{2024^{29}+2024}\)

=>\(1-\dfrac{2023}{2024^{30}+2024}>1-\dfrac{2023}{2024^{29}+2024}\)

=>\(\dfrac{A}{2024}>\dfrac{B}{2024}\)

=>A>B

Diện tích hình chữ nhật là \(345\cdot295=101775\left(m^2\right)\)

Diện tích hình chữ nhật là:

345 × 295 = 101775 (m²)

\(P=\dfrac{200}{2}+\dfrac{200}{6}+...+\dfrac{200}{9900}\)

\(=200\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{9900}\right)\)

\(=200\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)

\(=200\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=200\left(1-\dfrac{1}{100}\right)=200\cdot\dfrac{99}{100}=198\)

A) Những chữ cái có tâm đối xứng là: S, I, O, N

B) Những chữ cái có tâm đối xứng là: A, B, C, D, I, O, T, E, M, H

 Đường thẳng bất kì tạo với 2018 đường còn lại 2018 giao điểm.

Có 2019 đường như vậy nên ta có: 2018.2019 giao điểm.

 Nhưng mỗi giao điểm được tính 2 lần nên thực tế số giao điểm là: 2037171 giao điểm.

Đặt 2^x+2^x+1+...+2^x+2015=A

⇒ 2A=2^x+1+2^x+2...+2^x+2016

⇒ 2A-A= (2^x+1+2^x+2...+2^x+2016)-(2^x+2^x+1+...+2^x+2015)

⇒ A= 2^x+2016-2^x

⇒ 2^x+2016-2^x=2²⁰¹⁹-8=2²⁰¹⁹-2³

⇒x=3

Vậy x=3

\(\dfrac{15}{37}\cdot\dfrac{8}{27}-\dfrac{15}{37}\cdot\dfrac{37}{27}+3033\cdot\dfrac{15}{37}\)

\(=\dfrac{15}{37}\left(\dfrac{8}{27}-\dfrac{37}{27}+3033\right)\)

\(=\dfrac{15}{37}\cdot\left(-\dfrac{29}{27}+3033\right)\)

\(=\dfrac{15}{37}\cdot\dfrac{81862}{27}=\dfrac{81862}{37}\cdot\dfrac{5}{9}=\dfrac{409310}{333}\)

ĐKXĐ: n<>3

Để \(\dfrac{n+1}{n-3}\) là số nguyên thì \(n+1⋮n-3\)

=>\(n-3+4⋮n-3\)

=>\(4⋮n-3\)

=>\(n-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{4;2;5;1;7;-1\right\}\)

đkxd n ≠ 3