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\(\left(\dfrac{1}{3}\right)^2.\dfrac{14}{11}-\dfrac{3}{11}\left(\dfrac{1}{3}\right)^2\\ =\dfrac{1}{9}.\dfrac{14}{11}-\dfrac{3}{11}.\dfrac{1}{9}\\ =\dfrac{1}{9}\left(\dfrac{14}{11}-\dfrac{3}{11}\right)\\ =\dfrac{1}{9}.\dfrac{11}{11}\\ =\dfrac{1}{9}.1\\ =\dfrac{1}{9`}\)
a) = 0,6 - 0,8 = -0,2
b) 1/5 + |-4/5|
= 1/5 + 4/5
= 1
c) 3/7 . (-1/9) + 3/7 . (-2/3) - 5/3
= 3/7 . (-1/9 - 2/3) - 5/3
= 3/7 . 7/9 - 5/3
= 1/3 - 5/3
= -4/3
\(a,\left(\dfrac{3}{8}-\dfrac{1}{5}-\dfrac{1}{3}\right)+\left(\dfrac{5}{8}-1\dfrac{3}{7}-\dfrac{4}{7}\right)+\dfrac{1}{3}\\ =\dfrac{3}{8}-\dfrac{1}{5}-\dfrac{1}{3}+\dfrac{5}{8}-\dfrac{10}{7}-\dfrac{4}{7}\\ =\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{10}{7}+\dfrac{4}{7}\right)\\ =1-\dfrac{1}{5}-2=\left(1-2\right)-\dfrac{1}{5}\\ =-1-\dfrac{1}{5}=-\dfrac{6}{5}\)
\(b,-24\left(\dfrac{1}{3}-\dfrac{1}{4}\right)^2-1,5\\ =-24\left(\dfrac{4}{12}-\dfrac{3}{12}\right)^2-\dfrac{3}{2}\\ =-24\left(\dfrac{1}{12}\right)^2-\dfrac{3}{2}\\ =-24\cdot\dfrac{1}{144}-\dfrac{3}{2}\\ =-\dfrac{2}{37}-\dfrac{3}{2}=\dfrac{4}{74}-\dfrac{111}{74}\\ =\dfrac{-115}{74}\)
4 6 .256 2 .2 4 = 2 2 6 . 2 8 2 .2 4 = 2 12 .2 16 .2 4 = 2 32
c: Ta có: \(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=1-1+\dfrac{1}{64}\)
\(=\dfrac{1}{64}\)
93*92+14*16
\(=4\left(93\cdot23+14\cdot4\right)\)
\(=4\cdot2195=8780\)
\(A=-\dfrac{1}{6}-\left[\dfrac{9}{16}-\dfrac{9}{8}+\dfrac{3}{4}\right]\)
\(=\dfrac{-1}{6}-\dfrac{9-18+12}{16}=\dfrac{-1}{6}-\dfrac{3}{16}\)
\(=\dfrac{-8}{48}-\dfrac{9}{48}=-\dfrac{17}{48}\)
\(B=\left(-\dfrac{3}{4}+\dfrac{3}{4}\right)^2+\dfrac{3}{4}\cdot\dfrac{-5}{9}+7^3\)
=-15/36+343
=343-5/12=4111/12
\(\dfrac{7}{2}-\left(\dfrac{3}{4}+\dfrac{1}{5}\right)\\ =\dfrac{7}{2}-\left(\dfrac{15}{20}+\dfrac{4}{20}\right)\\ =\dfrac{7}{2}-\dfrac{19}{20}\\ =\dfrac{70}{20}-\dfrac{19}{20}\\ =\dfrac{51}{20}\)