a) \(\dfrac{25}{6}:\dfrac{5}{3}-\left(\dfrac{-1}{4}\right)\)

\(=\dfrac{25}{6}\cdot\dfrac{3}{5}+\dfrac{1}{4}\)

\(=\dfrac{5}{2}+\dfrac{1}{4}\)

\(=\dfrac{10}{4}+\dfrac{1}{4}\)

\(=\dfrac{11}{4}\)

b) \(\dfrac{1}{8}\cdot\dfrac{3}{2}+\dfrac{1}{8}\cdot\dfrac{-1}{2}\)

\(=\dfrac{1}{8}\cdot\left(\dfrac{3}{2}-\dfrac{1}{2}\right)\)

\(=\dfrac{1}{8}\cdot\dfrac{2}{2}\)

\(=\dfrac{1}{8}\)

a) 256:53−(−14)625​:35​−(4−1​)

=256⋅35+14=625​⋅53​+41​

=52+14=25​+41​

=5.2+14=45.2+1​

=114=411​

b) 18⋅32+18⋅(−12)81​⋅23​+81​⋅(−21​)

=18⋅(32−12)=81​⋅(23​−21​)

=18⋅22=81​⋅22​

=18⋅1=18=81​⋅1=81​
 

A = 2023 + 2023² + 2023³ + ... + 2023¹⁰⁰⁰⁰

⇒ 2023A = 2023² + 2023³ + 2023⁴ + ... + 2023¹⁰⁰⁰¹

⇒ 2022A = 2023A - A

= (2023² + 2023³ + 2023⁴ + ... + 2023¹⁰⁰⁰¹) - (2023 + 2023² + 2023³ + ... + 2023¹⁰⁰⁰⁰) 

= 2023¹⁰⁰⁰¹ - 2023

⇒ 2022A + 2023 = 2023¹⁰⁰⁰¹ - 2023 + 2023

= 2023¹⁰⁰⁰¹

= 2023².2023⁹⁹⁹⁹ ⋮ 2023²

Vậy (2022A + 2023) ⋮ 2023²

Ngày thứ nhất đọc được :

   90 x 4 / 9 = 40 (trang)

Ngày thứ hai đọc được :

  90 x 2/5 = 36 (trang)

Ngày thứ ba đọc được :

  90 - 36 - 40 = 14 (trang)

\(-\dfrac{9}{16}.\dfrac{13}{3}-\left(-\dfrac{3}{4}\right)^2.\dfrac{19}{3}\\ =-\dfrac{9}{16}.\dfrac{13}{3}-\dfrac{9}{16}.\dfrac{19}{3}\\ =-\dfrac{9}{16}.\left(\dfrac{13}{3}+\dfrac{19}{3}\right)\\ =-\dfrac{9}{16}.\dfrac{32}{3}\\ =-\dfrac{3.3.16.2}{16.3}=-3.2\\ =-6\)

Điểm B ở đâu ra thế em?

3x+7 chia hết cho x

Nhận thấy : Với mọi số nguyên x thì 3x luôn chia hết cho x

Do đó để 3x+7 chia hết cho x

Thì : 7 phải chia hết cho x

=> \(x\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

cảm ơn nha

\(A=3+3^2+3^3+...+3^{199}+3^{200}\\ \Rightarrow3A=3^2+3^3+3^4+....+3^{200}+3^{201}\\ \Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{200}+3^{201}\right)-\left(3+3^2+3^3+...+3^{199}+3^{200}\right)\\ \Rightarrow2A=3^{201}-3\\ \Rightarrow2A+3=3^{201}\)( Là một lũy thừa của 3 ) => DPCM

cảm ơn

\(a.\dfrac{31}{17}+\left(-\dfrac{5}{13}\right)+\left(-\dfrac{8}{13}\right)-\dfrac{4}{17}\\ =\left(\dfrac{31}{17}-\dfrac{4}{17}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\\ =\dfrac{27}{17}+\left(-1\right)\\ =\dfrac{27}{17}+\dfrac{-17}{17}=\dfrac{10}{17}\\ b.\left(-2\right)^3-1=-8-1=-9\\ \dfrac{5}{27}\cdot\left(\dfrac{-3}{2}\right)^3=\dfrac{5}{27}\cdot\dfrac{-27}{8}=-\dfrac{5}{8}\)

a: \(\dfrac{3}{5}+\dfrac{3}{5\cdot9}+...+\dfrac{3}{97\cdot101}\)

\(=\dfrac{3}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{97\cdot101}\right)\)

\(=\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{4}\cdot\dfrac{100}{101}=\dfrac{75}{101}\)

b: \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{33}\right)=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

c: \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)

d: \(\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\cdot...\cdot\left(1-\dfrac{2014}{7}\right)\)

\(=\left(1-\dfrac{7}{7}\right)\cdot\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot...\cdot\dfrac{-2007}{7}\)

\(=\left(1-1\right)\cdot\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot...\cdot\dfrac{-2007}{7}\)

=0

\(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

=>\(x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\)

=>\(x+\dfrac{8}{45}=\dfrac{29}{45}\)

=>\(x=\dfrac{29-8}{45}=\dfrac{21}{45}=\dfrac{7}{15}\)