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\(125\%\cdot\left(-\frac{1}{2}\right)^2\div\left(1\frac{5}{6}-1,5\right)+2016^0\)
\(=\frac{125}{100}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{4}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{9}{6}\right)+1\)
\(=\frac{5}{16}\div\frac{2}{6}+1\)
\(=\frac{15}{16}+1=\frac{31}{16}\)
\(125\%.\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1,5\right)+2016^0\)
\(=\frac{125}{100}.\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}:\left(\frac{11}{6}-\frac{9}{6}\right)+1\)
\(=\frac{5}{16} :\frac{1}{3}+1\)
\(=\frac{5}{16} .\frac{3}{1}+1\)
\(=\frac{15}{16}+1\)
\(=\frac{31}{16}\)
1/ Bg
\(\frac{21^4}{27.\left(-343\right)}\)= \(\frac{\left(3.7\right)^4}{3^3.\left(-7\right)^3}\)
= \(\frac{3^4.7^4}{3^3.\left(-7\right)^3}\)
= \(\frac{3.\left(-7\right)}{1.1}\)
= 3.(-7)
= -21
2/ Bg
Ta có: \(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}\)và a + b - c = 21 (a, b, c thuộc Z)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}=\frac{a+b-c}{4+5-2}=\frac{21}{7}\)= 3
=> a = 3.4 = 12
=> b = 3.5 = 15
=> c = 3.2 = 6
Vậy a = 12, b = 15 và c = 6
= [ 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
= -4/3 - 17/6 . 6/11 + 3 : 1/20
= - 4/3 - 17/11 + 60
= 1885/33
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
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
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 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)
= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)
= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)
= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)
= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)
= \(-\frac{38}{15}+\frac{1}{23}\)
= \(-\frac{859}{345}\)
-20
-20