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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: Số số hạng là:
(40-2):2+1=20(số)
Tổng là:
42x10=420
\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)
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) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
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) \(=-\dfrac{80}{9}\)
a) 23.15 -[115-(12-5)2 ]
= 23.15 -[115-36]
= 8.15 -79
= 120-79
=41
b)5.[(85 - 35 : 7) :8 + 90 ] - 50
=5 .[80:8+90]-50
=5.100-50
=500-50
=450
c){[261 - ( 36-31)3.2 ]-9}.1001
={[261 - 125.2 ]-9}.1001
={[261 -250 ]-9}.1001
={11-9}.1001
=2.1001
=2002
d)3.102 - [1200 - ( 42 - 2.3)3]
=3.100-[1200 -(16-6)3 ]
=300-[1200-1000]
=300-200
=100
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) \(\left(\dfrac{3}{2}-\dfrac{2}{3}\right)+\dfrac{7}{6}\)
\(=\left(\dfrac{9}{6}-\dfrac{4}{6}\right)+\dfrac{7}{6}\)
\(=\dfrac{13}{6}+\dfrac{7}{6}\)
\(=\dfrac{20}{6}\)
\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)