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ta có \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}\)
= \(\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
= \(\frac{3^{29}.2^3}{2^2.3^{28}}=3.2=6\)
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
Bài 1:
\(\left(\frac{9}{4}:\frac{6}{5}\right)-1=\frac{45}{24}-1=\frac{15}{8}-1=\frac{7}{8}\)
\(2+\left(\frac{1}{4}\times\frac{2}{5}\right)=2+\frac{2}{10}=2+\frac{1}{5}=\frac{11}{5}\)
Bài 2:
\(\frac{3}{5}+\frac{2}{3}=\frac{9}{15}+\frac{10}{15}=\frac{19}{15}\)
\(1-\frac{9}{11}=\frac{1}{1}-\frac{9}{11}=\frac{11}{11}-\frac{9}{11}=\frac{2}{11}\)
\(\frac{7}{9}\times\frac{3}{14}=\frac{21}{126}=\frac{3}{18}\)
\(\frac{15}{7}:\frac{5}{21}=\frac{315}{35}=9\)
1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+9+8+9+10+12+13+14+15+16+17+18+19+20+30+1000000=10000274
\(\frac{14}{15}-\frac{5}{6}=\frac{1}{10}\)
\(\frac{9}{7}-\frac{9}{14}=\frac{9}{14}\)
\(2-\frac{3}{4}=\frac{5}{4}\)
c) \(\left(11+12+13+...+19\right).\left(6.8-48\right)=\left(11+12+13+...+19\right).\left(48-48\right)\)
\(=\left(11+12+13+...+19\right).0\)
\(=0\)
Hok "tuốt" nha^^
b) \(\left(500.9-250.18\right).\left(1+2+3+...+9\right)=\left(250.2.9-250.18\right).\left(1+2+...+9\right)\)
\(=\left(250.18-250.18\right).\left(1+2+3+...+9\right)\)
\(=0.\left(1+2+3+...+9\right)\)
\(=0\)
Hok tốt nha^^ (lần 2)
Toán lớp 6 má đưa zô đây làm cái zề ???