Quả dưa hấu nặng:

\(\dfrac{15}{4}:\dfrac{3}{4}=5\left(kg\right)\)

Đáp số: 5kg 

 $\genfrac{}{}{0ex}{}{15}{4}:\genfrac{}{}{0ex}{}{3}{4}=5$ (kg)

Ta có:

\(\dfrac{-2}{3}< \dfrac{-1}{3}< 0< \dfrac{1}{2}\)

Sắp xếp:
\(\dfrac{-2}{3};\dfrac{-1}{3};0;\dfrac{1}{2}\)

−23;−13;0;12

3

−2

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;

3

−1

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;0;

2

1

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.

$\genfrac{}{}{0ex}{}{-2}{3};\genfrac{}{}{0ex}{}{-1}{3};0;\genfrac{}{}{0ex}{}{1}{2}$.

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) $\genfrac{}{}{0ex}{}{25}{6}:\genfrac{}{}{0ex}{}{5}{3}-\left(\genfrac{}{}{0ex}{}{-1}{4}\right)$

$=\genfrac{}{}{0ex}{}{25}{6}\cdot \genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{1}{4}$

$=\genfrac{}{}{0ex}{}{5}{2}+\genfrac{}{}{0ex}{}{1}{4}$

$=\genfrac{}{}{0ex}{}{5.2+1}{4}$

$=\genfrac{}{}{0ex}{}{11}{4}$

b) $\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{3}{2}+\genfrac{}{}{0ex}{}{1}{8}\cdot \left(-\genfrac{}{}{0ex}{}{1}{2}\right)$

$=\genfrac{}{}{0ex}{}{1}{8}\cdot \left(\genfrac{}{}{0ex}{}{3}{2}-\genfrac{}{}{0ex}{}{1}{2}\right)$

$=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{2}{2}$

$=\genfrac{}{}{0ex}{}{1}{8}\cdot 1=\genfrac{}{}{0ex}{}{1}{8}$
 

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

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