Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
(22+21+22+23).20.21.22.23
=(4+2+4+8).1.2.4.8
=18.1.2.4.8
=1152
1 3/8+1/8:(0,75-1/2)-25%.1/2
=11/8+1/8:(3/4-1/2)-1/4.1/2
=12/8:1/4-1/8
=6/1-1/8
=47/8
12 1/3-5/6:(24-23 5/7)
=37/3-5/6:(24-166/7)
=37/3-5/6:2/7
=37/3-35/2
=31/6
(-1/2)2-(-2)2-50
=1/4-4-1
=-19/4
\(\left(2^2+2^1+2^2+2^3\right)×2^0×2^1×2^2×2^3\)
\(=\left(4+2+4+8\right)×1×2×4×8\)
\(=18×1×2×4×8\)
\(=1152\)
\(1\frac{3}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%×\frac{1}{2}\)
\(=\frac{11}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}×\frac{1}{2}\)
\(=\frac{11}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
\(=\frac{11}{8}+\frac{1}{2}-\frac{1}{8}\)
\(=\frac{7}{4}\)
\(12\frac{1}{3}-\frac{5}{6}:\left(24-23\frac{5}{7}\right)\)
\(=\frac{37}{3}-\frac{5}{6}:\left(24-\frac{166}{7}\right)\)
\(=\frac{37}{3}-\frac{5}{6}:\frac{2}{7}\)
\(=\frac{37}{3}-\frac{35}{12}\)
\(=\frac{113}{12}\)
\(\left(\frac{-1}{2}\right)^2-\left(-2\right)^2-5^0\)
\(=\frac{1}{4}-4-1\)
\(=\frac{-19}{4}\)
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
\(C=\frac{8\cdot5-8\cdot2}{16}=\frac{8\cdot(5-2)}{16}=\frac{1\cdot(5-2)}{2}=\frac{1\cdot3}{2}=\frac{3}{2}\)
\(H=\left[\frac{-5}{24}+0,75+\frac{7}{12}\right]:\left[-2\frac{1}{8}\right]\)
\(H=\left[\frac{-5}{24}+\frac{75}{100}+\frac{7}{12}\right]:\left[-\frac{17}{8}\right]\)
\(H=\frac{9}{8}:\left[\frac{-17}{8}\right]=\frac{9}{8}\cdot\frac{8}{-17}=\frac{9}{1}\cdot\frac{1}{-17}=\frac{9}{-17}=\frac{-9}{17}\)