\(\dfrac{x}{7}+\dfrac{2}{y}=\dfrac{-1}{15}\)

=>\(\dfrac{xy+14}{7y}=\dfrac{-1}{15}\)

=>\(15\left(xy+14\right)+7y=0\)

=>\(15xy+7y=-210\)

=>y(15x+7)=-210

mà 15x+7 chia 15 dư 7

nên \(\left(15x+7;y\right)\in\left\{\left(7;-30\right)\right\}\)

=>\(\left(x;y\right)\in\left(0;-30\right)\)

help me pls:(

a) \(x-\dfrac{3}{5}=\dfrac{2}{7}\)

\(\Rightarrow x=\dfrac{2}{7}+\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{10}{35}+\dfrac{21}{35}\)

\(\Rightarrow x=\dfrac{31}{35}\)

b) \(x+\dfrac{20}{11\cdot13}+\dfrac{20}{13\cdot15}+...+\dfrac{20}{53\cdot55}=\dfrac{3}{11}\)

\(\Rightarrow x+10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)

\(\Rightarrow x+10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Rightarrow x+10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(\Rightarrow x+10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)

\(\Rightarrow x+\dfrac{40}{55}=\dfrac{3}{11}\)

\(\Rightarrow x=\dfrac{3}{11}-\dfrac{40}{55}\)

\(\Rightarrow x=\dfrac{-25}{55}=\dfrac{-5}{11}\)

a) 

\(\dfrac{4}{5}-\left(\dfrac{-2}{3}\right)-\dfrac{1}{10}-\dfrac{2}{3}\\ =\dfrac{4}{5}+\dfrac{2}{3}-\dfrac{1}{10}-\dfrac{2}{3}\\ =\left(\dfrac{4}{5}-\dfrac{1}{10}\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)\\ =\dfrac{7}{10}+0\\ =\dfrac{7}{10}\)  

b) 

\(\dfrac{1}{3}-\dfrac{-1}{2}+\dfrac{1}{13}-\dfrac{5}{6}\\ =\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\dfrac{1}{13}-\dfrac{5}{6}\\ =\dfrac{5}{6}+\dfrac{1}{13}-\dfrac{5}{6}\\ =\dfrac{1}{13}\) 

c) 

\(\dfrac{-5}{12}-\left(\dfrac{-5}{6}-\dfrac{5}{12}\right)\\ =\dfrac{-5}{12}+\dfrac{5}{6}+\dfrac{5}{12}\\ =\left(-\dfrac{5}{12}+\dfrac{5}{12}\right)+\dfrac{5}{6}\\ =\dfrac{5}{6}\)

Ta có :

\(\dfrac{1300}{1500}=\dfrac{13}{15}=1-\dfrac{2}{15}\)

\(\dfrac{1333}{1555}=1-\dfrac{222}{1555}\)

Vì \(\dfrac{222}{1555}>\dfrac{2}{15}\)

\(\Rightarrow1-\dfrac{222}{1555}< 1-\dfrac{2}{15}\)

\(\dfrac{\Rightarrow1333}{1555}< \dfrac{1300}{1500}\)

hi

\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2008}-1\right).\left(\dfrac{1}{2009}-1\right)\)

\(=-\dfrac{1}{2}.-\dfrac{2}{3}.-\dfrac{3}{4}...-\dfrac{2007}{2008}.-\dfrac{2008}{2009}\)

\(=-\left(\dfrac{1.2.3...2007.2008}{2.3.4...2008.2009}\right)\)

\(=-\dfrac{1}{2009}\)

4(x-2)-3(x+1)=5

=>\(4x-8-3x-3=5\)

=>\(x-11=5\)

=>x=11+5=16

\(4\left(x-2\right)-3\left(x+1\right)=5\)

\(\Leftrightarrow4x-8-3x-3=5\)

\(\Leftrightarrow\left(4x-3x\right)=5+8+3\)

\(\Leftrightarrow x=16\)

Vậy \(x=16\)

Ta có:

\(\dfrac{1}{2}=\dfrac{1\times2}{2\times2}=\dfrac{2}{4};\dfrac{1}{4}=\dfrac{1}{4}\)

Vì \(\dfrac{2}{4}>\dfrac{1}{4}\) nên \(\dfrac{1}{2}>\dfrac{1}{4}\)

Mọi người ơi, cuộc thi Bài văn số 326 đâu rồi nhỉ? 

= > Cậu vô olm.vn rồi lướt xuống là sẽ thấy.

\(\dfrac{1}{2}>\dfrac{1}{4}\left(\dfrac{1}{4}< \dfrac{1}{2}\right)\)

\(\dfrac{29}{2}=\dfrac{28+1}{2}=14+\dfrac{1}{2}=14\dfrac{1}{2}\)

\(\dfrac{15}{4}=\dfrac{12+3}{4}=3+\dfrac{3}{4}=3\dfrac{3}{4}\)

\(\dfrac{31}{2}=\dfrac{30+1}{2}=15\dfrac{1}{2}\)

\(\dfrac{29}{3}=\dfrac{27+2}{3}=9\dfrac{2}{3}\)

\(\dfrac{125}{8}=\dfrac{120+5}{8}=15+\dfrac{5}{8}=15\dfrac{5}{8}\)

\(\dfrac{36}{27}=\dfrac{27+9}{27}=1+\dfrac{9}{27}=1\dfrac{9}{27}\)

\(\dfrac{124}{15}=\dfrac{120+4}{15}=8+\dfrac{4}{15}=8\dfrac{4}{15}\)

\(\dfrac{96}{3}=\dfrac{93+3}{3}=31\dfrac{3}{3}\)

\(\dfrac{129}{24}=\dfrac{120+9}{24}=5+\dfrac{9}{24}=5\dfrac{9}{24}\)

\(\dfrac{78}{13}=\dfrac{65+13}{13}=5+\dfrac{13}{13}=5\dfrac{13}{13}\)

\(\dfrac{91}{4}=\dfrac{88+3}{4}=22+\dfrac{3}{4}=22\dfrac{3}{4}\)

\(\dfrac{115}{8}=\dfrac{112+3}{8}=14+\dfrac{3}{8}=14\dfrac{3}{8}\)

a: \(B=2021\times2025=\left(2023-2\right)\times\left(2023+2\right)=2023\times2023-2\times2\)

=>\(B=A-4\)

=>A lớn hơn B 4 đơn vị

b: \(C=35\times53-18=35\times35+35\times18-18\)

\(=35\times35+18\times\left(35-1\right)\)

\(=35\times35+18\times34\)

\(D=35+53\times34\)

\(=35+\left(35-1\right)\times\left(35+18\right)\)

\(=35+35\times35+35\times18-35\times1-18\)

\(=35\times35+35\times17+17=35\times35+36\times17\)

\(=35\times35+18\times34\)

=C

=>C=D

Gọi số cần tìm có dạng là \(X=\overline{ab}\)

Vì viết thêm số 7 vào bên trái số đó thì sẽ được số mới gấp 36 lần số cần tìm nên ta có: \(\overline{7ab}=36\times\overline{ab}\)

=>\(700+\overline{ab}=36\times\overline{ab}\)

=>\(35\times X=700\)

=>X=20

Vậy: Số cần tìm là 20