23\(\dfrac{1}{4}\).\(\dfrac{7}{5}\)-13\(\dfrac{1}{4}\):\(\dfrac{5}{7}\)
4) \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
= \(\dfrac{93}{4}.\dfrac{7}{5}-\dfrac{53}{4}:\dfrac{5}{7}\)
= \(\dfrac{93}{4}.\dfrac{7}{5}-\dfrac{53}{4}.\dfrac{7}{5}\)
= \(\dfrac{7}{5}.\left(\dfrac{93}{4}-\dfrac{53}{4}\right)\)
= \(\dfrac{7}{5}.10\)
= \(14\)
1) \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
2) \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(0,8-\dfrac{3}{4}\right)^2\)
3) \(16\dfrac{2}{7}:\left(\dfrac{-3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
4) \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
5) \(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4\)
1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
a)\(\dfrac{7}{8}\)x\(\dfrac{3}{13}\)+\(\dfrac{4}{9}\)x\(\dfrac{4}{13}\)
b)\(\dfrac{6}{5}\)+\(\dfrac{7}{3}\)+\(\dfrac{8}{9}\)
c)23: \(\dfrac{5}{14}\)+\(\dfrac{6}{7}\)+\(\dfrac{4}{9}\)
d)\(4\dfrac{1}{4}\)+\(7\dfrac{3}{7}\)-\(2\dfrac{4}{17}\)
e)8-(9\(\dfrac{2}{11}\)+\(\dfrac{8}{33}\))
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
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: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
1)\(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
2)\(\dfrac{ }{\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}}\)
3)\(\dfrac{ }{\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}}\)
4)\(\dfrac{ }{\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}}\)
1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{19}{18}+\dfrac{5}{18}\)
\(=\dfrac{24}{18}\)
\(=\dfrac{4}{3}\)
2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\dfrac{1}{15}+\dfrac{7}{15}\)
\(=\dfrac{8}{15}\)
3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)
(1,5 điểm) Tính giá trị của biểu thức.
a) $\dfrac{7}{2}-\left(\dfrac{3}{4}+\dfrac{1}{5}\right)$
b) $\dfrac{12}{23} . \dfrac{7}{13}+\dfrac{11}{23} . \dfrac{7}{13};$
c) $|-2|-\left(\dfrac{5}{9}-\dfrac{2}{3}\right)^2: \dfrac{4}{27}.$
a) 7/2 - (3/4 + 1/5)
= 7/2 - 19/20
= 51/20
b) 12/23 . 7/13 + 11/23 . 7/13
= 7/13 . (12/23 + 11/23)
= 7/13 . 1
= 7/13
c) |-2| - (5/9 - 2/3)² : 4/27
= 2 - 1/81 : 4/27
= 2 - 1/12
= 23/12
a) 7/2 - (3/4 + 1/5)
= 7/2 - 19/20
= 51/20
b) 12/23 . 7/13 + 11/23 . 7/13
= 7/13 . (12/23 + 11/23)
= 7/13 . 1
= 7/13
c) |-2| - (5/9 - 2/3)² : 4/27
= 2 - 1/81 : 4/27
= 2 - 1/12
= 23/12
a) \(\dfrac{7}{2}\)-(\(\dfrac{3}{4}\)+\(\dfrac{1}{5}\))
=\(\dfrac{7}{2}\)-(\(\dfrac{15}{20}\)+\(\dfrac{4}{20}\))
=\(\dfrac{7}{2}\)-\(\dfrac{19}{20}\)
=\(\dfrac{70}{20}\)-\(\dfrac{19}{20}\)
\(\dfrac{51}{20}\)
b) \(\dfrac{12}{23}\).\(\dfrac{7}{13}\)+\(\dfrac{11}{23}\).\(\dfrac{7}{13}\)
=\(\dfrac{7}{13}\).(\(\dfrac{12}{23}\)+\(\dfrac{11}{23}\))
=\(\dfrac{7}{13}\).1
=\(\dfrac{7}{13}\)
c) ∣−2∣-(\(\dfrac{5}{9}\)-\(\dfrac{2}{3}\))\(^2\):\(\dfrac{4}{27}\)
= 2-(\(\dfrac{-1}{9}\))\(^2\):\(\dfrac{4}{27}\)
= 2-\(\dfrac{1}{81}\):\(\dfrac{4}{27}\)
= 2-\(\dfrac{1}{12}\)
\(\dfrac{23}{12}\)
Giá trị x thoả mãn \(2\dfrac{1}{4}x - 6\dfrac{3}{5} = 3,75 \) là :
A. \(\dfrac{4}{5}\) B. \(\dfrac{23}{5}\)
C.\(\dfrac{13}{5}\) D. \(\dfrac{1}{7}\)
Bài 1.( 2 điểm)Tính bằng cách hợp lí:
a) \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
b) \(\left(\dfrac{18}{23}+\dfrac{7}{12}\right)+\left(\dfrac{-13}{19}-\dfrac{3}{4}\right)+\left(\dfrac{-6}{19}+\dfrac{5}{23}\right)\)
c) \(\dfrac{4}{3}+\dfrac{-5}{6}+\dfrac{-1}{4}\)
d) \(\dfrac{5}{6}-\dfrac{7}{5}+\dfrac{17}{30}\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
Bài 2: tính
a, \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
b,\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
c,\(\dfrac{7}{23}.[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}]\)
d,\(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
e,\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-1728\)
c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)
\(=\dfrac{7}{23}.\dfrac{-23}{6}\)
\(=\dfrac{-7}{6}\)
d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)
\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)
\(=10.\dfrac{7}{5}\)
\(=14\)
e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
Tính
a, 4\(\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
b, \(\dfrac{4^6\times9^5+6^9\times120}{-8^4\times3^{12}+6^{11}}\)
c, \(\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{12}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{10}}\)
d,\(\dfrac{30\times4^7\times3^{29}-5\times14^5\times2^{12}}{54\times6^{14}\times9^7-12\times8^5\times7^5}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0