tổng dfrac{1}{2} + dfrac{-3}{4} bằng:A.dfrac{1}{4} B.-1C.dfrac{-5}{4}D.1
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tổng \(\dfrac{1}{2}\) + \(\dfrac{-3}{4}\) bằng:
A.\(\dfrac{1}{4}\)
B.-1
C.\(\dfrac{-5}{4}\)
D.1
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Kết quả giới hạn \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2}{x-1}\) bằng:
A. 0
B. \(\dfrac{1}{2}\)
C. \(\dfrac{1}{4}\)
D. \(\dfrac{1}{3}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{\sqrt{1+3}+2}=\dfrac{1}{4}\)
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20 tháng 1 2023 lúc 14:23
Độ dài cung 300 của một đường tròn bán kính 4 cm bằng:
A.\(\dfrac{4}{3}\pi cm\) B.\(\dfrac{2}{3}\pi cm\) C.\(\dfrac{1}{3}\pi cm\) D.\(\dfrac{8}{3}\pi cm\)
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Áp dụng công thức :
\(l=\dfrac{\pi Rn}{180}=\dfrac{\pi.4.30^o}{180^o}=\dfrac{2}{3}\pi cm\\ =>B\)
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Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\)
c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)
d) 40% \(-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\)
\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)
\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)
\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)
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a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)
\(=\dfrac{2}{5}+-2\)
\(=\dfrac{2}{5}+\dfrac{-10}{5}\)
\(=\dfrac{-8}{5}\)
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a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{5}+-2=\dfrac{-8}{5}\)
b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)=\dfrac{19}{5}-\dfrac{9}{4}-\dfrac{9}{5}=\left(\dfrac{19}{5}-\dfrac{9}{5}\right)-\dfrac{9}{4}=2-\dfrac{9}{4}=\dfrac{-1}{4}\)
c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)
\(=\dfrac{-3}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\)
\(=\dfrac{-3}{5}.1+\dfrac{3}{5}\)
\(=\dfrac{-3}{5}+\dfrac{3}{5}\)
\(=0\)
d) \(40\%-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\)
\(=\dfrac{2}{5}-\dfrac{12}{7}:\dfrac{3}{7}+\dfrac{9}{5}\)
\(=\dfrac{2}{5}-4+\dfrac{9}{5}\)
\(=\dfrac{-9}{5}\)
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3. Tính :a/ dfrac{-1}{2} + dfrac{5}{6} + dfrac{1}{3} b/ dfrac{-3}{8} + dfrac{7}{4} - dfrac{1}{12} c/ dfrac{3}{5} : (dfrac{1}{4} . dfrac{7}{5}) d/ dfrac{10}{11} + dfrac{4}{11} : 4 - dfrac{1}{8}
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3. Tính :
a/ \(\dfrac{-1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{1}{3}\) b/ \(\dfrac{-3}{8}\) + \(\dfrac{7}{4}\) - \(\dfrac{1}{12}\) c/ \(\dfrac{3}{5}\) : (\(\dfrac{1}{4}\) . \(\dfrac{7}{5}\)) d/ \(\dfrac{10}{11}\) + \(\dfrac{4}{11}\) : 4 - \(\dfrac{1}{8}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
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a,dfrac{2}{3}xdfrac{5}{2}:dfrac{9}{5}b,dfrac{1}{3}xdfrac{1}{4}+dfrac{5}{6}c,dfrac{1}{2}-dfrac{7}{8}:dfrac{7}{4}d,dfrac{6}{5}-dfrac{4}{5}xdfrac{3}{2}
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a,\(\dfrac{2}{3}\)x\(\dfrac{5}{2}\):\(\dfrac{9}{5}\)
b,\(\dfrac{1}{3}\)x\(\dfrac{1}{4}\)+\(\dfrac{5}{6}\)
c,\(\dfrac{1}{2}\)-\(\dfrac{7}{8}\):\(\dfrac{7}{4}\)
d,\(\dfrac{6}{5}\)-\(\dfrac{4}{5}\)x\(\dfrac{3}{2}\)
a, \(\dfrac{1}{3}-\dfrac{1}{4}:\dfrac{2}{5}\)
b, \(\dfrac{6}{7}-\left(\dfrac{5}{6}+\dfrac{1}{3}\right)-\left(\dfrac{2}{3}+\dfrac{1}{7}\right)\)
c, \(\dfrac{-5}{9}.\dfrac{2}{5}+4\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{-3}{5}\)
d, \(3\dfrac{1}{2}-\left(5\dfrac{4}{7}-1\dfrac{1}{2}\right):0,75\)
a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`
b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`
`=(6/7-1/7)-(1/3+2/3)-5/6`
`=5/7-1-5/6`
`=-47/42`
c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`
`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`
`=-5/9 . (2/5 + 3/5-1) + 4`
`=-5/9 . 0 +4`
`=4`
d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`
`=7/2 - (39/7 - 3/2) : 3/4`
`= 7/2 - 57/14 : 3/4`
`=7/2 - 38/7`
`=-27/14`
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Tính:
a) \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\);
b) \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\);
c) \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)\);
d) \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\).
a: \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)
\(=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}\)
\(=\dfrac{4}{6}=\dfrac{2}{3}\)
b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)
\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}\)
\(=\dfrac{31}{24}\)
c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)=\dfrac{3}{4}:\dfrac{7}{20}=\dfrac{3}{4}\cdot\dfrac{20}{7}=\dfrac{15}{7}\)
d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)
\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{7}{8}\)
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a,dfrac{8^{20}+4^{20}}{4^{25}+64^5}b,left(1+dfrac{2}{3}-dfrac{1}{4}right).left(dfrac{4}{5}-dfrac{3}{4}right)^2c,23dfrac{1}{3}:left(dfrac{-5}{7}right)-13dfrac{1}{3}:left(dfrac{-5}{7}right)d,1:left(dfrac{2}{3}-dfrac{3}{4}right)^2e,dfrac{45^{10}.5^{20}}{75^{15}}
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a,\(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}\)
b,\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
c,\(23\dfrac{1}{3}:\left(\dfrac{-5}{7}\right)-13\dfrac{1}{3}:\left(\dfrac{-5}{7}\right)\)
d,1:\(\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
e,\(\dfrac{45^{10}.5^{20}}{75^{15}}\)
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
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