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a: \(=\dfrac{6\left(12-7\right)}{60}=\dfrac{6\cdot5}{6\cdot10}=\dfrac{5}{10}=\dfrac{1}{2}\)
b: \(=\dfrac{35\cdot\left(18-1\right)}{34\cdot7}=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\)
c: \(=\dfrac{42\left(1-11\right)}{21\cdot\left(-15\right)}=2\cdot\dfrac{-10}{-15}=2\cdot\dfrac{2}{3}=\dfrac{4}{3}\)
d: \(=\dfrac{2^5\cdot3^2}{2^5\cdot3^3}=\dfrac{1}{3}\)
e: \(=\dfrac{-36\cdot14}{27\left(14+7\right)}=\dfrac{-36}{27}\cdot\dfrac{14}{21}=\dfrac{-4}{3}\cdot\dfrac{2}{3}=-\dfrac{8}{9}\)
\(\left(49\cdot42-47\cdot42\right):42\)
\(=\left[\left(49-47\right)\cdot42\right]:42\)
\(=2\cdot42:42\)
\(=2\)
khi bỏ dấu ngoặc biểu thức 120+(-8)-(42-57) ta được:
A.120+8-42-57
B.120-8-42-57
C.120-8-42+57
D.120+8-42+57
\(\left(39\times42-37\times42\right):42\)
\(=\left[42\left(39-37\right)\right]:42\)
\(=42.2:42=84:42\)
\(=2\)
\(=\left(\dfrac{19}{42}+\dfrac{23}{42}\right)+\left(\dfrac{1}{37}+\dfrac{36}{37}\right)+10\)
=12
\(-\dfrac{18}{42}\cdot\dfrac{35}{42}:\dfrac{-98}{42}\)
\(=-\dfrac{18}{42}\cdot\dfrac{35}{42}\cdot\dfrac{42}{-98}\)
\(=\dfrac{-18}{42}\cdot\dfrac{-5}{14}\\ =\dfrac{15}{98}\)
\(\dfrac{-18}{42}.\dfrac{3\text{5}}{42}:\dfrac{-98}{42}\)
\(=\dfrac{-18}{42}.\dfrac{3\text{5}}{42}.\dfrac{-42}{98}\)
\(=\dfrac{1\text{5}}{98}\)
42 . x = 39 . 42 - 37. 42
=> 42x=42.(39-37)
=> 42x=42.2
=> x=2
( 39 . 42 - 37 . 42) : 42
= [ 42 . ( 39 - 37) ] : 42
= [ 42 . 2 ] : 42
= 84 : 42
= 2
^^ Cbht!!!
Ta có : 42.88+11.42+42
= 42.(88 + 11 + 1)
= 42 . 100
= 4200
42.88+11.42+42
=42.(88+11+1)
=42.100
=4200