\(\sqrt{\dfrac{9}{4}}-\sqrt{2}+\sqrt{2}\\ =\dfrac{3}{2}-\left(\sqrt{2}-\sqrt{2}\right)\\ =\dfrac{3}{2}-0\\ =\dfrac{3}{2}\)

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}\\ =-\dfrac{1}{3}\)

\(\left(7-\dfrac{1}{5}+\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\\ =7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\\ =\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)\\ =1-2-1\\ =-2\)

\(1-\left(4\cdot\dfrac{2}{5}+x-7\cdot\dfrac{2}{3}\right):15\cdot\dfrac{1}{3}=0\\ \left(\dfrac{8}{5}+x-\dfrac{14}{3}\right):15\cdot\dfrac{1}{3}=1\\ \left(\dfrac{8}{5}+x-\dfrac{14}{3}\right):15=1:\dfrac{1}{3}\\ \left(\dfrac{8}{5}+x-\dfrac{14}{3}\right):15=3\\ \dfrac{8}{5}+x-\dfrac{14}{3}=3\cdot15\\ x-\dfrac{46}{15}=45\\ x=45+\dfrac{46}{15}\\ x=\dfrac{629}{15}\)

Vậy...

Số chia là:

\(44:8=5\) (dư 4)

Đáp số: 5

Em đăng kí nhận bằng tiền mặt

Em đăng kí nhận thưởng của câu lạc bộ chiến binh OLM tháng 6 năm 2024

nên:

\(x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\\\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}-10=0\\ \left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0 \\ \dfrac{x-85-15}{15}+\dfrac{x-74-26}{13}+\dfrac{x-67-33}{11}+\dfrac{x-64-36}{9}=0\\ \dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\\ \left(x-100\right)\cdot\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)

Vì \(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\ne0\)

a) Sửa đề: (n+7) chia hết cho (n+3)

\(\left(n+7\right)⋮\left(n+3\right)\\ \Rightarrow\left(n+3+4\right)⋮\left(n+3\right)\)

\(\Rightarrow\)\(4⋮\left(n+3\right)\)

Mà \(n\) là số tự nhiên nên \(n+3\) cũng là số tự nhiên suy ra:

\(\left(n+3\right)\inƯ\left(4\right)=\left\{1,2,4\right\}\)

\(\Rightarrow n\in\left\{-2,-1,1\right\}\)

\(\Rightarrow n=1\) (thỏa mãn điều kiện)

Vậy...

b)

 \(\left(2n+9\right)⋮\left(n+1\right)\\\Rightarrow \left(2n+2+7\right)⋮\left(n+1\right)\\ \left[2\left(n+1\right)+7\right]⋮\left(n+1\right)\\ 7⋮\left(n+1\right)\)

Mà \(n\) là số tự nhiên nên \(n+1\) cũng là số tự nhiên suy ra:

\(\left(n+1\right)\inƯ\left(7\right)=\left\{1,7\right\}\)

\(\Rightarrow n\in\left\{0,6\right\}\) (thỏa mãn điều kiện)

Vậy...