Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{3}{5}.2,75-1,2:\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}:\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}.\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(\dfrac{-2}{5}+\dfrac{8}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{11}{4}.0\)
\(=0\)
\(-2\dfrac{3}{4}\) (-0,4) + \(1\dfrac{2}{5}\) . 2,75 - (-1,2) : \(\dfrac{4}{11}\)
\(B=\dfrac{\left(81,624:4\dfrac{4}{3}-4,505\right)^2+125\dfrac{3}{4}}{\left\{\left[\left(\dfrac{11}{25}\right)^2:0,88+3,53\right]^2-\left(2,75\right)^2\right\}.\dfrac{13}{25}}\)
\(B=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1^1_6-0,875+0,7}\right).\dfrac{1842009}{1842010}\)
\(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
THỰC HIỆN PHÉP TÍNH NHA MN
-11 \(\dfrac{3}{4}\) - ( \(6\dfrac{5}{6}\) - \(4\dfrac{1}{2}\) ) - \(1\dfrac{2}{3}\)
1, so sánh A;B biết: A=\(\left(\dfrac{\left(3\cdot\dfrac{2}{15}+\dfrac{1}{5}\right):2\cdot\dfrac{1}{2}}{\left(5\cdot\dfrac{3}{7}-2\cdot\dfrac{1}{4}\right):\dfrac{443}{56}}\right);B=\dfrac{1,2:\left(1\cdot\dfrac{1}{5}.1\cdot\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
Chứng minh rằng P>3 biet P= \(\dfrac{5}{2×1}+\dfrac{4}{1×11}+\dfrac{3}{11×2}+\dfrac{1}{2×15}+\dfrac{13}{15×4}+\dfrac{15}{4×43}+\dfrac{13}{43×8}\)
cho A=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
CMR:0,2<A<0,4
\(\left(\dfrac{2}{3}+\dfrac{2}{13}+\dfrac{2}{11}+\dfrac{2}{6}\right)\) \(\left(\dfrac{5}{4}+\dfrac{5}{7}+\dfrac{5}{6}+\dfrac{5}{11}\right)\)
___________________ X ___________________
\(\left(\dfrac{4}{3}+\dfrac{4}{13}+\dfrac{4}{11}+\dfrac{4}{6}\right)\) \(\left(\dfrac{9}{4}+\dfrac{9}{7}+\dfrac{9}{6}\dfrac{9}{11}\right)\)
