1 + 4 = 5 2 + 7 = 12 3 + 6 = 21 8 + 11 = ??
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a.\(\dfrac{27}{8}\)
b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)
d.\(\dfrac{7}{3}\)
e.5
g.\(\dfrac{53}{16}\)
Bài 1 :
a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)
b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)
c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)
d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)
e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)
g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)
1. a
\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)
\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)
\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)
1.b
\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)
\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)
2.a
\(\dfrac{4}{5}+x=\dfrac{11}{10}\)
\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)
2.b
\(x-\dfrac{3}{4}=\dfrac{5}{7}\)
\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)
Lời giải:
a.
$\frac{7}{4}+\frac{5}{-6}+\frac{21}{8}=\frac{42}{24}+\frac{-20}{24}+\frac{63}{24}=\frac{85}{24}$
b.
$\frac{4}{9}+\frac{-7}{12}+\frac{8}{15}$
$=\frac{80}{180}+\frac{-105}{180}+\frac{96}{180}=\frac{71}{180}$
c.
$=\frac{-1}{3}+\frac{5}{6}+\frac{19}{12}+2$
$=\frac{-4}{12}+\frac{10}{12}+\frac{19}{12}+2=\frac{25}{12}+2=\frac{49}{12}$
\(1-2-3+4+5-6-7+8+...+97-98-99+100+101\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(97-98-99+100\right)+101\)
\(=0+0+...+0+101\)
\(=101\)
1 + 4 = 5
52 + 7 = 59
123 + 6 = 129
218 + 11 = 229
1 + 4 = 5
52 + 7 = 59
123 + 6 = 129
218 + 11 = 229