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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) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
21)
\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\\ =\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{10000}{9999}\\ =\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{100.100}{99.101}\\ =\dfrac{2.3.4.....100}{1.2.3.....99}.\dfrac{2.3.4.....100}{3.4.5.....101}\\ =100.\dfrac{2}{101}\\ =\dfrac{200}{101}\)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right):\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right):\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Rightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right):6069=\frac{1}{3}\)
\(\Rightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Rightarrow1488x-476,5x=20230\)
\(\Rightarrow1011,5x=20230\)
\(\Rightarrow x=20\)
Bài làm :
Ta có :
\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right)\div\left(1+8+15+...+281+288\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right)\div\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right)\div6069=\frac{1}{3}\)
\(\Leftrightarrow\frac{1488x-279+279-476,5x}{10}=2023\)
\(\Leftrightarrow1488x-476,5x=20230\)
\(\Leftrightarrow1011,5x=20230\)
\(\Leftrightarrow x=20\)
Vậy x=20
B=\(\frac{1}{3}+\frac{-0,175+\frac{5}{11}}{0,2625-\frac{15}{22}}\)
B=\(\frac{1}{3}+\frac{\frac{123}{440}}{-\frac{369}{880}}\)
B=\(0,2161148416\)
Thay x vào ta có:
\(B=\frac{1}{3}+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\)
\(B=\frac{1}{3}+\frac{\frac{2}{10}-\frac{375}{1000}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)
\(B=\frac{1}{3}+\frac{\frac{1}{5}-\frac{3}{8}+\frac{5}{11}}{-\frac{3}{10}+\frac{9}{16}-\frac{15}{22}}\)
\(B=\frac{1}{3}+\frac{\frac{123}{440}}{\frac{-369}{880}}\)
\(B=\frac{1}{3}+\frac{123}{440}.\frac{880}{-369}\)
\(B=\frac{1}{3}+\frac{-2}{3}\)
\(B=\frac{-1}{3}\)
Vậy...