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Nguyễn Thị Ái Vân
Giới thiệu về bản thân
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a/ \(\dfrac{1}{5}+\dfrac{4}{5}:x=0,75\) b/ \(x+\dfrac{1}{2}=1-x\)
\(\dfrac{1}{5}+\dfrac{4}{5}:x=\dfrac{3}{4}\) \(\left(1-x\right)-x=\dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{4}{5}:x=\dfrac{3}{4}\) \(1-2x=\dfrac{1}{2}\)
\(\dfrac{4}{5}:x=\dfrac{3}{4}-\dfrac{1}{5}\) \(2x=1-\dfrac{1}{2}\)
\(\dfrac{4}{5}:x=\dfrac{15}{20}-\dfrac{4}{20}\) \(2x=\dfrac{2}{2}-\dfrac{1}{2}\)
\(\dfrac{4}{5}:x=\dfrac{11}{20}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{4}{5}:\dfrac{11}{20}\) \(x=\dfrac{1}{2}:2\)
\(x=\dfrac{4}{5}.\dfrac{20}{11}\) \(x=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{80}{55}=\dfrac{16}{11}\) \(x=\dfrac{1}{4}\)
a/ \(\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}\) b/ \(2.\left(\dfrac{-3}{2}\right)^2-\dfrac{7}{2} \) c/\(\left(-\dfrac{3}{4}\right).5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
\(=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)\) \(=\dfrac{2}{1}.\dfrac{9}{4}-\dfrac{7}{2}\) \(=\left(-\dfrac{3}{4}\right).\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
\(=\dfrac{2}{3}.\dfrac{1}{2}\) \(=\dfrac{9}{2}-\dfrac{7}{2}\) \(=\left(-\dfrac{3}{4}\right).\left(\dfrac{68}{13}+\dfrac{36}{13}\right)\)
\(=\dfrac{2.1}{3.2}\) \(=\dfrac{9-7}{2}\) \(=\left(-\dfrac{3}{4}\right).\dfrac{8}{1}\)
\(=\dfrac{2}{6}=\dfrac{1}{3}\) \(=\dfrac{2}{2}=1\) \(=\dfrac{\left(-3\right).8}{4.1}\)
\(=\dfrac{-24}{4}=-6\)
Bài 1.
a/ \(\left(-1,62\right)+\dfrac{2}{5}+x=7\) b/ \(4\dfrac{3}{5}-x=\dfrac{-1}{5}+\dfrac{1}{2}\) c/ \(\dfrac{-4}{7}-x=\dfrac{3}{5}-2x\) d/ \(\dfrac{5}{7}-\dfrac{1}{13}+0,25=3\dfrac{1}{2}-x\)
\(\left[\left(-1,62\right)+0,4\right]+x=7\) \(\dfrac{23}{5}-x=\dfrac{3}{10}\) \(2x-x=\dfrac{3}{5}-\dfrac{-4}{7}\) \(\dfrac{5}{7}-\dfrac{1}{13}+\dfrac{1}{4}=\dfrac{7}{2}-x\)
\(\left(-1,22\right)+x=7\) \(-x=\dfrac{3}{10}-\dfrac{23}{5}\) \(2x-x=\dfrac{41}{35}\) \(\dfrac{29}{49}+\dfrac{1}{4}=\dfrac{7}{2}-x\)
\(x=7+1,22\) \(-x=-\dfrac{43}{10}\) \(x=\dfrac{41}{35}\) \(\dfrac{165}{196}\) \(=\dfrac{7}{2}-x\)
\(x=8,22\) \(x=\dfrac{43}{10}\) \(-x=\dfrac{165}{196}-\dfrac{7}{2}\)
\(-x=-\dfrac{521}{196}\)
\(x=\dfrac{521}{196}\)
Bài 1.
a/ \(A=\left(\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}\right)+\left(\dfrac{2}{3}+\dfrac{-7}{5}+1\dfrac{1}{7}\right)\)
\(=\dfrac{1}{3}-\dfrac{8}{15}-\dfrac{1}{7}+\dfrac{2}{3}+\dfrac{-7}{5}+\dfrac{8}{7}\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(-\dfrac{1}{7}+\dfrac{8}{7}\right)+\left(-\dfrac{8}{15}+\dfrac{-7}{5}\right)\)
\(=\) \(1+\left(-1\right)+\dfrac{-29}{15}\)
\(=0+\dfrac{-29}{15}\)
\(=\dfrac{-29}{15}\)
Trò chơi | Số bạn chọn |
Cướp cờ | 5 |
Kéo co | 9 |
Nhảy bao bố | 12 |
Bịt mắt bắt dê | 8 |
Ô ăn quan | 6 |
a/ \(\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{5}{6}\) b/\(\dfrac{-2}{3}+\dfrac{6}{5}:\dfrac{2}{3}-\dfrac{2}{15}\) c/\(\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}\) d/\(\dfrac{12}{19}+\dfrac{-8}{13}-\dfrac{12}{19}+\dfrac{5}{-13}+2\)
= \(\dfrac{4}{12}+\dfrac{9}{12}-\dfrac{10}{12}\) = \(\dfrac{-2}{3}+\dfrac{9}{5}-\dfrac{2}{15}\) = \(\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\dfrac{5}{13}\) = \(\left[\dfrac{12}{19}+\left(-\dfrac{12}{9}\right)\right]+\left(\dfrac{-8}{13}+\dfrac{5}{-13}\right)+2\)
= \(\dfrac{13}{12}-\dfrac{10}{12}\) = \(\dfrac{-10}{15}+\dfrac{27}{15}-\dfrac{2}{15}\) = \(\left(-1\right)+\dfrac{5}{13}\) = \(0+\left(-1\right)+2\)
= \(\dfrac{13-10}{12}\) = \(\dfrac{-17}{15}-\dfrac{2}{15}\dfrac{ }{ }\) = \(\dfrac{-13}{13}+\dfrac{5}{13}\) = \(\left(-1\right)+2\)
= \(\dfrac{2}{12}=\dfrac{1}{6}\) = \(\dfrac{\left(-17\right)-2}{15}\) = \(\dfrac{\left(-13\right)+5}{13}\) = \(1\)
= \(\dfrac{19}{15}\) = \(\dfrac{-8}{13}\)