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a) A = 53 - 6 . 52 + 6 . 5 - 5
= 5 ( 52 - 6 . 5 + 6 - 1 )
= 5 ( 25 - 30 + 6 - 1 )
= 5 . 0
= 0
b) Câu b theo mình đề thế này nhé bạn
\(B=\frac{-3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)
\(B=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(B=-3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=-3\left(\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(B=-3\left(\frac{5}{4.5}-\frac{4}{4.5}+\frac{6}{5.6}-\frac{5}{5.6}+\frac{7}{6.7}-\frac{6}{6.7}+\frac{8}{7.8}-\frac{7}{7.8}+\frac{9}{8.9}-\frac{8}{8.9}+\frac{10}{9.10}-\frac{9}{9.10}\right)\)
\(B=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=-3\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(B=-3\left(\frac{5}{20}-\frac{2}{20}\right)\)
\(B=-3.\frac{3}{20}\)
\(B=\frac{-9}{20}\)
Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
Bài làm
a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)
= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)
= \(-\frac{3}{7}+\frac{7}{2}\)
\(=-\frac{7}{14}+\frac{49}{14}\)
\(=\frac{42}{14}=3\)
b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)
\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)
\(=5-\frac{1}{3}+\frac{2}{9}\)
\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)
\(=\frac{44}{9}\)
c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)
\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)
\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)
\(=\frac{5}{12}+\frac{7}{12}\)
\(=\frac{12}{12}=1\)
d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)
\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)
\(=-\frac{5}{9}+\frac{1}{90}\)
\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)