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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)
\(\Rightarrow x=50\)
Vậy x = 50
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)
\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)
\(\frac{25}{3}=8\frac{1}{3}\)
\(\frac{15+x}{5}=3\frac{x}{5}\)
vì 0 < x < 5 nên ko tách ra được hỗn số . kk
\(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.20}\right)-x+\frac{221}{231}=\frac{4}{3}\)
\(=\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{20}\right)-x=\frac{4}{3}-\frac{221}{231}\)
\(=\left(\frac{1}{11}-\frac{1}{20}\right)-x=\frac{29}{77}\)
\(=\frac{9}{220}-x=\frac{29}{77}\)
\(x=\frac{9}{220}-\frac{29}{77}\)
bạn ơi chỗ \(\frac{2}{19.20}\) có phải là \(\frac{2}{19.21}\) không
\(\frac{1}{5}=\frac{1.3}{5.3}=\frac{3}{15}\)
\(\frac{-10}{55}=\frac{-10\div5}{55\div5}=\frac{-2}{11}\)
Vậy ba cặp số phân số bằng nhau sau khi sử dụng tính chất cơ bản
2 .
\(\frac{-12}{-3}=\frac{-12:3}{-3:3}=\frac{-4}{-1};\frac{7}{-35}=\frac{7:7}{-35:7}=\frac{1}{-5};\frac{-9}{27}=\frac{-9:9}{27:9}=\frac{-1}{3}\)
3 .
\(15min=\frac{1}{4}\)giờ
\(90min=\frac{3}{2}\)giờ
a) Ta có: \(2 = \frac{2}{1} = \frac{{2.15}}{{1.15}} = \frac{{30}}{{15}} < \frac{{31}}{{15}}\).
Suy ra \(\frac{{31}}{{15}} > 2\).
b) Ta có: \( - 3 = \frac{{ - 3}}{1} = \frac{{ - 3.2}}{{1.2}} = \frac{{ - 6}}{2}\)
và \(\frac{7}{{ - 2}} = \frac{{ - 7}}{2}\)
Do \(\frac{{ - 6}}{2} > \frac{{ - 7}}{2}\) nên \( - 3 > \frac{7}{{ - 2}}\).
\(a,\frac{6\cdot9-2\cdot18}{53\cdot3-126}=\frac{2\cdot3\cdot3\cdot3-2\cdot3\cdot3\cdot2}{53\cdot3-53\cdot2}\)
\(=\frac{2\cdot3^3-2^2\cdot3^2}{53\cdot3-53\cdot2}=\frac{2\cdot3^2\cdot\left(3-2\right)}{53\cdot\left(3-2\right)}\)
\(=\frac{2\cdot9}{53}=\frac{18}{53}\)
\(b,\frac{21\cdot17-5\cdot17}{19\cdot20-40}=\frac{21\cdot17-5\cdot17}{19\cdot20-2\cdot20}\)
\(=\frac{17\left(21-5\right)}{20\cdot\left(19-2\right)}=\frac{17\cdot16}{20\cdot17}\)
\(=\frac{16}{20}=\frac{4}{5}\)
a, \(\frac{6.9-2.18}{53.3-126}=\frac{2.3.3.3-2.3.3.2}{53.3-53.2}\)
\(=\frac{2.3^3-2^2.3^2}{53.3-53.2}=\frac{2.3^2.\left(3-2\right)}{53.\left(3-2\right)}\)
\(=\frac{2.9}{53}=\frac{18}{53}\)
b, \(\frac{21.17-6.17}{19.20-40}=\frac{21.17-6.17}{19.20-2.20}\)
\(=\frac{17\left(21-6\right)}{20\left(19-2\right)}=\frac{17.15}{20.17}\)
\(=\frac{15}{20}=\frac{3}{4}\)
Study well ! >_<
ko biết