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\(\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}=\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\)
\(\frac{1}{1}-\frac{1}{6}=\frac{5}{6}\)
k nha gõ mỏi tay lắm
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\cdot\left(2x+3\right)}\)
\(=\frac{1}{3}\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{\left(2x+1\right)\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x+3}{3\left(2x+3\right)}-\frac{3}{3\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x+3-3}{3\left(2x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{2x}{6x+9}\right)\)
\(=\frac{2x}{3\left(6x+9\right)}=\frac{2x}{18x+27}\)
https://olm.vn/hoi-dap/question/893332.html
câu trả lời đó
mk thấy nó cx giống vs bài của cậu
hok tốt
\(2\frac{4}{5}.3\frac{1}{8}\) \(1\frac{1}{5}:1\frac{4}{5}\)
= \(\frac{14}{5}.\frac{25}{8}\) = \(\frac{6}{5}:\frac{9}{5}\)
= \(\frac{35}{4}\) = \(\frac{2}{3}\)
tk mk nha
\(2\frac{4}{5}\cdot3\frac{1}{8}\)
\(=\frac{14}{5}\cdot\frac{25}{8}\)
\(=\frac{35}{4}\)
\(1\frac{1}{5}:1\frac{4}{5}\)
\(=\frac{6}{5}:\frac{9}{5}\)
\(=\frac{2}{3}\)
a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)
\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
(1-1/2) x (1-1/3) x (1-1/4) x (1-1/5)
=1/2*2/3*3/4*4/5
=1/5.
1-1/2 x 1-1/3 x 1-1/4 x 1-1/5
=> = \(\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{5}=\frac{1}{120}\)
đúng rồi nha > - <