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\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)
\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)
\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)
\(x=\dfrac{198}{99}\)
Vậy \(x=\dfrac{198}{99}\)
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)
và cứ như thế đến số cuối
Theo đề: \(2x+y=0\Leftrightarrow y=-2x\) \(\left(1\right)\)
Ta có:
\(\dfrac{3-x}{y-4}=\dfrac{2}{5}\)
\(\Leftrightarrow5\left(3-x\right)=2\left(y-4\right)\)
\(\Leftrightarrow15-5x=2y-8\)
\(\Leftrightarrow15+8=2y+5x\)
\(\Leftrightarrow5x+2y=23\) \(\left(2\right)\)
Thế (1) vào (2), suy ra:
\(5x+2.\left(-2x\right)=23\)
\(\Leftrightarrow5x-4x=23\)
\(\Leftrightarrow x=23\)
\(\Rightarrow y=-2.23=-46\)
A bn lướt xuống dưới mà xem cách làm
nhưng của bn là cho 3 ra ngoài nha
\(\dfrac{-5}{6}=\dfrac{-5\times4}{6\times4}=\dfrac{-20}{24}\)
\(\dfrac{3}{-8}=\dfrac{3\times\left(-3\right)}{-8\times\left(-3\right)}=\dfrac{-9}{24}\)
\(2=\dfrac{48}{24}\)
\(\dfrac{-25}{100}=\dfrac{-1}{4}=\dfrac{-1\times6}{4\times6}=\dfrac{-6}{24}\)
\(\dfrac{72}{108}=\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)
Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2013.2015}\)
\(\Leftrightarrow2A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}\right)\)
\(\Leftrightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2015}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2015}=\dfrac{2012}{6045}\)
\(\Leftrightarrow A=\dfrac{1006}{6045}\)
2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{2013.2015}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}+\dfrac{1}{2015}\)
2A=\(\dfrac{1}{1}-\dfrac{1}{2015}\)
2A=\(\dfrac{2014}{2015}\)
A=\(\dfrac{1007}{2015}\)
Khi gặp bài này, bn nên tách 1 phân số ra thành hiệu của 2 phân số.
\(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)-x=\dfrac{-100}{99}\\ \left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=\dfrac{-100}{99}\\ \left(1-\dfrac{1}{99}\right)-x=\dfrac{-100}{99}\\ \dfrac{98}{99}-x=\dfrac{-100}{99}\\ x=\dfrac{98}{99}-\dfrac{-100}{99}\\ x=\dfrac{198}{99}\\ x=2\)
\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=\dfrac{-100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)\(-x=\dfrac{-100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=\dfrac{-100}{99}\)
\(\dfrac{98}{99}-x=\dfrac{-100}{99}\)
\(x=\dfrac{98}{99}-\dfrac{-100}{99}\)
\(x=2\)