a: \(\dfrac{-2}{9}+\dfrac{8}{15}+1\dfrac{2}{3}-\dfrac{7}{9}+\dfrac{-7}{-15}\)

\(=\left(-\dfrac{2}{9}-\dfrac{7}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{5}{3}\)

\(=-1+1+\dfrac{5}{3}=\dfrac{5}{3}\)

b: \(\dfrac{2}{7}\cdot10\dfrac{2}{8}-\dfrac{2}{7}\cdot3\dfrac{1}{4}\)

\(=\dfrac{2}{7}\left(10+\dfrac{1}{4}-3-\dfrac{1}{4}\right)\)

\(=\dfrac{2}{7}\cdot7=2\)

c: \(\dfrac{5}{7}+\dfrac{2}{7}:0,25-0,4\)

\(=\dfrac{5}{7}+\dfrac{2}{7}\cdot4-\dfrac{2}{5}\)

\(=\dfrac{5}{7}+\dfrac{8}{7}-\dfrac{2}{5}=\dfrac{13}{7}-\dfrac{2}{5}=\dfrac{65-14}{35}=\dfrac{51}{35}\)

\(B=\dfrac{n+1}{n-2}=\dfrac{n-2+3}{n-2}=\dfrac{n-2}{n-2}+\dfrac{3}{n-2}=1+\dfrac{3}{n-2}\)

Để B nguyên

\(\Rightarrow3⋮n-2\) 

\(\Rightarrow n-2\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

\(\Rightarrow n\in\left\{3;5;1;-1\right\}\)

Sửa đề: Tìm n nguyên để B nguyên

ĐKXĐ: n<>2

Để B nguyên thì \(n+1⋮n-2\)

=>\(n-2+3⋮n-2\)

=>\(3⋮n-2\)

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

=>\(n\in\left\{3;1;5;-1\right\}\)

\(D=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)

\(D=\dfrac{1}{49}+1+\dfrac{2}{48}+1+....+\dfrac{48}{2}+1+1\)

\(D=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)

\(D=50\left(\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}+\dfrac{1}{50}\right)\)

\(\Rightarrow\dfrac{C}{D}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

cám ơn

Bạn bổ sung đề bài chỗ "cứ ... m² thì thu được" là bao nhiêu nhé.

a: Số tiền ông Bách nhận được sau 1 năm là:

\(100000000\left(1+7,3\%\right)=107300000\left(đồng\right)\)

b: Số tiền ông Bách nhận được sau 2 năm là:

\(107300000\left(1+7,3\%\right)=115132900\left(đồng\right)\)

800 triệu cuối cùng chiếm:

\(1-\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{5}{12}\)(tổng số tiền)

Tổng số tiền là:

\(800000000:\dfrac{5}{12}=800\cdot10^6\cdot\dfrac{12}{5}=1920000000\left(đồng\right)\)

-108 ko thể được số nào mũ 2 được nhé vì âm nhân âm = dương và dương nhân dương= dương

a: \(\left(\dfrac{1}{4}+\dfrac{-3}{11}\right)+\left(\dfrac{2}{11}+\dfrac{-8}{11}+\dfrac{3}{4}\right)\)

\(=\dfrac{1}{4}+\dfrac{-3}{11}+\dfrac{2}{11}+\dfrac{-8}{11}+\dfrac{3}{4}\)

\(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)-\dfrac{9}{11}=1-\dfrac{9}{11}=\dfrac{2}{11}\)

b: \(\left(-1\dfrac{1}{3}+1\dfrac{1}{2}\right)+2\dfrac{1}{2}\)

\(=-1-\dfrac{1}{3}+1+\dfrac{1}{2}+2+\dfrac{1}{2}\)

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

c: \(\left(-2\dfrac{1}{4}\right)+\dfrac{5}{2}:3\dfrac{3}{4}\)

\(=-\dfrac{9}{4}+\dfrac{5}{2}:\dfrac{15}{4}\)

\(=-\dfrac{9}{4}+\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{-9}{4}+\dfrac{2}{3}\)

\(=\dfrac{-27+8}{12}=-\dfrac{19}{12}\)

d: \(\dfrac{5}{8}:0,25-\left(0,6-2\dfrac{1}{4}\right):2\dfrac{1}{5}\)

\(=\dfrac{5}{8}:\dfrac{1}{4}-\left(\dfrac{3}{5}-\dfrac{9}{4}\right):\dfrac{11}{5}\)

\(=\dfrac{5}{2}-\dfrac{12-45}{20}\cdot\dfrac{5}{11}\)

\(=\dfrac{5}{2}-\dfrac{5}{20}\cdot\dfrac{-33}{11}=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{13}{4}\)

e: \(1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot...\cdot1\dfrac{1}{999}\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}\)

\(=\dfrac{1000}{2}=500\)

f: \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)

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