Tính hợp lí: \(\left(-2\right)^3\left(\frac{3}{4}-0,25\right):\left(\frac{2}{\frac{1}{4}}-\frac{1}{\frac{1}{6}}\right)\)
Những câu hỏi liên quan
1. Tính hợp lí :
a) \(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)
b) 7 + \(\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
2. Tìm các sô nguyên x biết:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
Câu 1:
a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)
\(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)
\(=\frac{1}{2}-\frac{4}{3}\)
\(=-\frac{5}{6}\)
b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
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Câu 2:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
Vậy -1\(\le\)x<7
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Tính hợp lí
a) \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+1\frac{1}{3}\right)\)
b) \(0,25+\frac{2}{5}+\left(\frac{1}{9}-1\frac{2}{5}+\frac{5}{4}\right)\)
Thực hiện phép tính bằng cách hợp lí
a, \(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac{2}{3}\) b, \(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
Thực hiện phép tính (hợp lí nếu có thể):1,frac{-1}{3}-frac{-3}{5}-frac{1}{6}+frac{1}{43}-frac{-3}{7}+frac{-1}{2}-frac{1}{35}
2,left(-frac{1}{3}+frac{7}{13}right)-left(frac{-16}{24}+frac{6}{26}+frac{9}{13}right)3,frac{-7}{3}-left[frac{2}{5}-left(frac{1}{3}+frac{-5}{25}right)right]
4,left(2frac{1}{4}-3frac{1}{5}right)-left[frac{-3}{4}+left(frac{4}{5}-2019right)right]
Đọc tiếp
Thực hiện phép tính (hợp lí nếu có thể):
\(1,\frac{-1}{3}-\frac{-3}{5}-\frac{1}{6}+\frac{1}{43}-\frac{-3}{7}+\frac{-1}{2}-\frac{1}{35}\\ \\ 2,\left(-\frac{1}{3}+\frac{7}{13}\right)-\left(\frac{-16}{24}+\frac{6}{26}+\frac{9}{13}\right)\)\(3,\frac{-7}{3}-\left[\frac{2}{5}-\left(\frac{1}{3}+\frac{-5}{25}\right)\right]\\ 4,\left(2\frac{1}{4}-3\frac{1}{5}\right)-\left[\frac{-3}{4}+\left(\frac{4}{5}-2019\right)\right]\)
a)\(-119\frac{3}{4}.\frac{3}{21}-\frac{13}{21}:4+\frac{6}{42}.116\frac{1}{2}\)
b)\(0,25.\left(-1199\frac{1}{2}\right)-\frac{1}{4}.\left(-2004\frac{1}{3}\right)-\frac{5}{4}\)
c)\(1\frac{3}{4}:\left(\frac{-1}{2}+\frac{2}{3}-\frac{5}{6}\right)-\left(-1,75\right)+25\%:2\)
Tính hợp lý:
1.Tính bằng cách hợp lí :
a)\(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{5}{6}+\left(\frac{1}{5}+\frac{5}{9}\right):\frac{5}{6}\)
b)\(\frac{1}{3^2}-\left(\frac{1}{3}\right)^2.\left(\frac{-1}{3}\right)^2\)
Tính \(A=\left(0,25\right)^{-1}.\left(\frac{1}{4}\right)^{-2}.\left(\frac{4}{3}\right)^2.\left(\frac{5}{4}\right)^{-1}.\left(\frac{2}{3}\right)^{-3}\)
\(A=\left(0,25\right)^{-1}.\left(\frac{1}{4}\right)^{-2}.\left(\frac{4}{3}\right)^2.\left(\frac{5}{4}\right)^{-1}.\left(\frac{2}{3}\right)^{-3}\)
\(\Rightarrow A=4^1.4^2.\frac{16}{9}.\frac{4}{5}\frac{27}{8}\)
\(\Rightarrow A=\frac{64}{1}.\frac{16}{9}.\frac{4}{5}.\frac{27}{8}\)
\(\Rightarrow A=\frac{1536}{5}\)
Vậy \(A=\frac{1536}{5}\)
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Thực hiện phép tính:
A=\(\left(3\frac{1}{3}+2,5\right):\left(3\frac{1}{6}-4\frac{1}{5}\right)-\frac{11}{31}\)
B=\(\left(-6\right).10:\left[-0,25+\frac{1}{2}:\left(-2\right)\right]+1\frac{3}{4}\)
A \(=\frac{35}{6}.\left(\frac{-30}{31}\right)-\frac{11}{31}=\frac{-175}{31}-\frac{11}{31}=\frac{-186}{31}=-6\)
\(B=-60:\left(\frac{-1}{2}\right)+\frac{7}{4}=30+\frac{7}{4}=\frac{127}{4}\)
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Tính bằng cách hợp lí ( Nếu có thể )
a) \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}\)
b) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)\)
a)\(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{117}{12}-\frac{2}{12}=\frac{115}{12}\)
b)\(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)
a. \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)
b. \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-1\frac{1}{2}\)
= \(\frac{13}{4}.1-\frac{3}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}\)
c. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)
Giải :
a) \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)
b) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)
c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)
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