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1.
a)=1/3-[(-5/4)-5/8]
=1/3-(-15/8)=53/24
b)=5/9:(-3/22)+5/9:(-3/5)
=5/9*22/-3+5/9*5/-3=-110/27+-25/27=5
2
a)Ta có 339<340=920<1120<1121
nên 339<1121
b)Ta có /3,4-x/ lớn hơn hoặc bằng 0 Với mọi x thuộc R
=> -/3,4-x/ bé hơn hoặc bằng 0 Với mọi x thuộc R
=> 0,5-/3,4-x/ bé hơn hoặc bằng 0,5 Với mọi x thuộc R
Dấu = xảy ra khi 3,4-x=0
=>x=3,4
Vậy GTLN của A = 0,5 khi x=3,4
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
tính:
a.(-2/3+3/7) : 4/5 + (-1/3+4/7) : 4/5
b. 5/9 : (1/11-5/22) + 5/9 : (1/15 - 2/3)
Help Me, Please !
a.(-2/3+3/7) : 4/5 + (-1/3+4/7) : 4/5
= [(-2/3 + 3/7) + (-1/3 + 4/7)] : 4/5
= [(-2/3 + (-1/3) + (3/7 + 4/7)] : 4/5
= [-1 + 1] : 4/5
= 0 : 4/5
= 0
a) \(\left(\frac{-2}{3}+\frac{3}{7}\right).\frac{5}{4}+\left(\frac{-1}{3}+\frac{4}{7}\right).\frac{5}{4}\)
=\(\left(\frac{-2}{3}+\frac{-1}{3}+\frac{3}{7}+\frac{4}{7}\right).\frac{5}{4}\)
= \(0.\frac{5}{4}=0\)
b) \(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}+\frac{1}{15}-\frac{2}{3}\right)\)
=\(\frac{5}{9}:\frac{-81}{110}=\frac{-550}{729}\)
a) \(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(\frac{-2}{3}+\frac{3}{7}+\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(-1+1\right):\frac{4}{5}\)
\(=0:\frac{4}{5}\)
\(=0\)
b) \(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{-3}{22}+\frac{5}{9}:\frac{-3}{5}\)
\(=\frac{5}{9}.\frac{-22}{3}+\frac{5}{9}.\frac{-5}{3}\)
\(=\frac{5}{9}.\left(\frac{-22}{3}+\frac{-5}{3}\right)\)
\(=\frac{5}{9}.-9\)
\(=-5\)
\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)
b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)
c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
