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\(8\frac{7}{10}+2\frac{3}{4}=\frac{87}{10}+\frac{11}{4}=\frac{174}{20}+\frac{55}{20}=\frac{229}{20}\)
Bạn chỉ cần đưa về phân số xong tính bình thường. Muốn đổi từ hỗn số sang phân số, ta chỉ cần lấy phần nguyên nhân cho mẫu rồi cộng với tử là xong. Chứ bạn cứ hỏi mấy bài dễ như thế này thì k giỏi đc đâu!!!
\(a.\)\(1\frac{2}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+5\frac{3}{7}\)
\(=\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{3}\cdot\frac{3}{2}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(=\frac{4}{2}+\frac{38}{7}\)
\(=2+\frac{38}{7}\)
\(=\frac{14}{7}+\frac{38}{7}\)
\(=\frac{52}{7}\)
\(b.1\frac{1}{3}-1\frac{1}{4}:1\frac{1}{2}+2\frac{3}{4}\cdot3\frac{2}{3}\)
\(=\frac{4}{3}-\frac{5}{4}:\frac{3}{2}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{4}\cdot\frac{2}{3}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{6}+\frac{121}{12}\)
\(=\frac{16}{12}-\frac{10}{12}+\frac{121}{12}\)
\(=\frac{6}{12}+\frac{121}{12}\)
\(=\frac{127}{12}\)
\(c.7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{2}{5}\cdot\frac{2}{1}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{14}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{70}{15}-\frac{12}{15}-\frac{10}{15}\)
\(=\frac{58}{15}-\frac{10}{15}\)
\(=\frac{48}{15}=\frac{16}{5}\)
\(\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(2+\frac{38}{7}\)
\(\frac{52}{7}\)
ở dãy 1 thì số đứng sau bằng tổng hai số đứng trước
ta có 5 số tiếp theo la 40,74, 136,...
a) \(1284-16401:231+58\) x\(11=1284-71+638\)
\(=1851\)
b)\(\frac{7}{4}+1\frac{1}{2}\)x \(1\frac{3}{2}:\frac{1}{4}=\frac{7}{4}+15\)
\(=\frac{67}{4}\)
c)\(2\frac{1}{4}+1\frac{2}{5}+\frac{4}{20}\)\(=\frac{9}{4}+\frac{7}{5}+\frac{1}{5}\)
\(=\frac{77}{20}\)
d)\(\left(\frac{12}{5}+\frac{2}{3}-\frac{11}{4}\right):\frac{3}{4}=\frac{19}{60}:\frac{3}{4}\)
\(=\frac{19}{60}\)x \(\frac{4}{3}\)
\(=\frac{19}{45}\)
a ) \(1\frac{1}{2}+2\frac{1}{3}+3\frac{1}{6}-5\)
\(=\frac{3}{2}+\frac{7}{3}+\frac{19}{6}-\frac{5}{1}\)
\(=\frac{9}{6}+\frac{14}{6}+\frac{19}{6}-\frac{30}{6}\)
\(=\frac{23}{6}+\frac{19}{6}-\frac{30}{6}\)
\(=\frac{42}{6}-\frac{30}{6}\)
\(=\frac{12}{6}=2\)
b ) \(2\frac{2}{3}\times3\frac{3}{4}\div4\frac{4}{5}\)
\(=\frac{8}{3}\times\frac{15}{4}\div\frac{24}{5}\)
\(=\frac{120}{12}\div\frac{24}{5}\)
\(=\frac{120}{12}\times\frac{5}{24}\)
\(=\frac{600}{288}=\frac{25}{12}\)
c ) \(4\frac{1}{5}+5\frac{1}{3}-2\frac{2}{3}\times3\frac{1}{5}+\frac{9}{25}\div\frac{9}{20}\)
\(=\frac{21}{5}+\frac{16}{3}-\frac{8}{3}\times\frac{16}{5}+\frac{9}{25}\div\frac{9}{20}\)
\(=\frac{63}{15}+\frac{80}{15}-\frac{128}{15}+\frac{9}{25}\times\frac{20}{9}\)
\(=\frac{143}{15}-\frac{128}{15}+\frac{180}{225}\)
\(=\frac{15}{15}+\frac{12}{15}\)
\(=\frac{27}{15}=\frac{9}{5}\)
2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{1.50}{100}-\frac{1}{100}=\frac{50-1}{100}=\frac{49}{100}\)
\(\left(1+1\frac{1}{4}+1\frac{1}{2}+1\frac{3}{4}+2+2\frac{1}{4}+2\frac{1}{2}+2\frac{3}{4}+...+4\frac{3}{4}\right):23\)
= \(\left(\frac{2}{2}+\frac{5}{4}+\frac{3}{2}+\frac{7}{4}+\frac{4}{2}+\frac{9}{4}+\frac{5}{2}+\frac{11}{4}+...+\frac{19}{4}\right):23\)
= \(\left(\frac{4}{4}+\frac{5}{4}+\frac{6}{4}+\frac{7}{4}+\frac{8}{4}+\frac{9}{4}+\frac{10}{4}+\frac{11}{4}+...+\frac{19}{4}\right):23\)
= \(\left(\frac{4+5+6+7+8+9+10+11+...+19}{4}\right):23\)
= \(\left(\frac{\left(19-4\right):1+1x\left(19+4\right):2}{4}\right):23\)
= \(46:23\)
= \(2\)