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Theo đề ta có:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)
=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)
= -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)
= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)
= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)
= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)
= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)
= \(-\frac{38}{15}+\frac{1}{23}\)
= \(-\frac{859}{345}\)
trong tích trên có 1 thừa số như thế này:
\(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{125}\right)\)
=0
=> tích trên bằng 0
= [ 60/90 - ( 12/15 + 10/15) ] : 6/5
= ( 2/3 - 22/15 ) x 5/6
= ( 10/15 - 22/15 ) x 5/6
= -12/15 x 5/6
= -60/90
= -2/3
( 15/10 x 4/9 - ( 4/5 + 2/3 )) : 6/5
( 15/10 x 4/9 - ( 12/15 + 10/15 )) : 6/5
(30/45 - 66/45 ) : 6/5
-12/15 : 6/5 ( đã rút gọn -36/45 = -12/15 )
-2/3
k mk na <3
a)\(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}\cdot\frac{1}{3}=4+\frac{1}{27}=\frac{108}{27}+\frac{1}{27}=\frac{109}{27}\)
b)\(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[\left(-8\right)\cdot2\right]=8+3-16=-5\)
a/ \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=5-1+\frac{1}{27}=4+\frac{1}{27}=\frac{109}{27}\)
b/ \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^3:\frac{1}{2}\right]=8+3.1+\left[-8:\frac{1}{2}\right]=11+-16=-5\)
a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)
\(=8+3.1+4:\frac{1}{2}\)
\(=8+3+8=19\)
b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)
a, \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)
\(=\frac{1}{3}.1-\frac{2}{3}\)
\(=\frac{1}{3}-\frac{2}{3}\)
\(=\frac{-1}{3}\)
b, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)
\(=\left(6-\frac{14}{5}\right).\frac{25}{8}+\frac{11}{8}.4\)
\(=\frac{16}{5}.\frac{25}{8}+\frac{11}{2}\)
\(=10+\frac{11}{2}\)
\(=\frac{31}{2}\)
1/3×(3/5+2/5)-2/15×1/5
1/3×1-2/15×1/5
1/3-2/15×1/5
1/3-2/75
25/75-2/75
23/75
(6-14/5)×25/8-11/8:4/1
16/5×25/8-11/8:4/1
10/1-11/8:4/1
10/1-11/8×1/4
10/1-11/32
320/32-11/32
309/32
\(125\%\cdot\left(-\frac{1}{2}\right)^2\div\left(1\frac{5}{6}-1,5\right)+2016^0\)
\(=\frac{125}{100}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{4}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{9}{6}\right)+1\)
\(=\frac{5}{16}\div\frac{2}{6}+1\)
\(=\frac{15}{16}+1=\frac{31}{16}\)
\(125\%.\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1,5\right)+2016^0\)
\(=\frac{125}{100}.\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}:\left(\frac{11}{6}-\frac{9}{6}\right)+1\)
\(=\frac{5}{16} :\frac{1}{3}+1\)
\(=\frac{5}{16} .\frac{3}{1}+1\)
\(=\frac{15}{16}+1\)
\(=\frac{31}{16}\)