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a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
a: =152,3+7,7+2021,19-2021,19
=160
b: =7/15*3/14*20/13
\(=\dfrac{7}{14}\cdot\dfrac{3}{15}\cdot\dfrac{20}{13}=\dfrac{1}{2}\cdot\dfrac{1}{5}\cdot\dfrac{20}{13}=\dfrac{2}{13}\)
c: \(=\dfrac{7}{4}\left(\dfrac{13}{12}-\dfrac{10}{12}\right)+\dfrac{5}{6}=\dfrac{7}{16}+\dfrac{5}{6}=\dfrac{61}{48}\)
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
A = \(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
A = \(-1\frac{1}{5}.\)4 : \(\frac{4.\left(1-\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5.\left(1-\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
A = \(-1\frac{1}{5}.4\): \(\frac{4}{5}\)= \(\frac{-6}{5}\).4. \(\frac{5}{4}\)
A = \(\frac{-24}{5}.\frac{5}{4}\)=\(\frac{\left(-6\right).1}{1.1}\)= -6.
\(A=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(=-1\frac{1}{5}.\frac{4}{1}:\frac{4}{5}\)
\(=-1\frac{1}{5}.\frac{4}{1}.\frac{5}{4}\)
\(=-1\)
a)[(-15).8]:4
=-120:4
=30
b)[(-125):(-5)].(-13)
=25.(-13)
=325
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
a) \(\left(\dfrac{1}{6}+\dfrac{5}{9}\right)+\dfrac{4}{9}\)
\(=\dfrac{1}{6}+\dfrac{5}{9}+\dfrac{4}{9}\)
\(=\dfrac{1}{6}+1\)
\(=\dfrac{7}{6}\)
b) \(\dfrac{3}{17}+\left(\dfrac{14}{17}-\dfrac{2}{3}\right)\)
\(=\dfrac{3}{17}+\dfrac{14}{17}-\dfrac{2}{3}\)
\(=1-\dfrac{2}{3}\)
\(=\dfrac{1}{3}\)
c) \(\left(\dfrac{3}{2}-\dfrac{2}{3}\right)+\dfrac{7}{6}\)
\(=\left(\dfrac{9}{6}-\dfrac{4}{6}\right)+\dfrac{7}{6}\)
\(=\dfrac{13}{6}+\dfrac{7}{6}\)
\(=\dfrac{20}{6}\)
A= 1 + (-3) + 5 +(-7) + ... +17 + (-19)
A= -10
nhớ k nha
A = 1 + (-3) + 5 + (-7) + ... + 17 + (-19)
A = [1 + (-3)] + [5 + (-7)] + ... + [17 + (-19)]
A = (-2) + (-2) + ... + (-2) Vì dãy A có 10 số => Dãy A có 5 cặp -2.
A = (-2) . 5
A = -10