Tính hợp lí nếu có thể
1/9 x 4/11 + 3/4 + 7/11 x 1/9
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= -5 / 7 x ( 2 / 11 + 9 / 11) + 12/7
= -5/7 x 1 + 12/7
= -5/7 + 12/7
= 1
b, = ( 9/4 x 105 - 9/4 x 101) : 3/2 - 64 x ( 1/4 - 3/4)
= [ 9/4 x ( 105 - 101)] : 3/2 -64 x -1/2
= 9/4 x 4 : 3 /2 - (-32)
= 9 : 3/2 + 32
= 18/3 +32
=
ta có
(12-12) +(10-9)+(8-7)+(5-4)+(2-1)+(3+11)
=0+1+1+1+1+14=18
12 - 12 +11 + 10 - 9 + 8 - 7 + 5 - 4 + 3 - 2 - 1
= ( 11 - 1 ) + ( 10 -9 ) + ( 8-7) + ( 5 - 4 ) + ( 3-2 )
= 10 + 1 + 1 + 1 + 1
=14
12-12+11+10-9+8-7+5-4+3-2-1
=(12-12)+(11+10)-(9-8)-(7-5)-(4-3)-(2+1)
=0+21-1-2-1-3
=21-(1+2+1+3)
=21-7
=14
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
1/9 x 4/11 + 3/4 +7/11 x 1/9
= 1/9 x (4/11 + 7/11) + 3/4
= 1/9 x 1 + 3/4
=1/9 + 3/4
= 31/36