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55 - 50 +45 - 40 + 35 -30 + 25 - 20 +15 - 10 +5
= 5 + 5 + 5 + 5 + 5 + 5
=5 x 6
=30
\(
\frac{48.700-24.45.20}{45-40+35-30+25-20+15-10+5}\)
\(=\frac{48.700-24.2.10.45}{5+5+5+5+5}\)
\(=\frac{48.700-48.450}{5.5}\)
\(=\frac{48.\left(700-450\right)}{25}\)
\(=\frac{48.250}{25}\)
\(=480\)
a)27*47+13*24+24*10*3
=27*47+13*24+24*30
=27*47+24*(13+30)
=27*47+24*43
=2301
b)
55-50+45-40+35-30+25-20+15-10+5
=(55-50)+(45-40)+(35-30)+(25-20)+(15-10)+5
=5+5+5+5+5+5
=30
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+1+\frac{36}{30}+\frac{49}{35}+\frac{64}{40}+\frac{81}{45}\)
\(=\left(\frac{1}{5}+\frac{81}{45}\right)+\left(\frac{4}{10}+\frac{49}{35}\right)+\left(\frac{9}{15}+\frac{49}{35}\right)+\left(\frac{16}{20}+\frac{36}{30}\right)+1\)
\(=2+2+2+2+1\)
\(=2\times4+1\)
\(=9\)
~ Hok tốt ~
Bài làm
\(A=\frac{0,24\cdot450+0,8\cdot15\cdot5+3\cdot3\cdot8}{65-60+55-50+45-40+35-30+25-20+5}\)
\(A=\frac{0,24\cdot45\cdot10+0,8\cdot15\cdot5+3\cdot3\cdot8}{\left(65-60\right)+\left(55-50\right)+\left(45-40\right)+\left(35-30\right)+\left(25-20\right)+5}\)
\(A=\frac{2,4\cdot45+0,8\cdot75+72}{5+5+5+5+5+5}\)
\(A=\frac{2,4\cdot45+0,8\cdot\frac{5}{3}\cdot45+1,6\cdot45}{30}\)
\(A=\frac{45\left(2,4+\frac{4}{5}\cdot\frac{5}{3}+1,6\right)}{30}\)
\(A=\frac{45\cdot\left(\frac{20}{15}+\frac{36}{15}+\frac{24}{15}\right)}{30}\)
\(A=\frac{3\cdot\frac{80}{15}}{2}\)
\(A=\frac{\frac{80}{5}}{2}\)
\(A=\frac{80}{5}\cdot\frac{1}{2}\)
\(A=\frac{40}{5}\)
\(A=8\)
1/5 + 4/10 +9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
=1/5 + 2/5 + 3/5 +...+ 9/5
=45/5=9
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+\frac{25}{25}+\frac{36}{30}+\frac{49}{35}+\frac{64}{40}+\frac{81}{45}\)
\(=\frac{1}{5}+\frac{1}{5}+\frac{3}{5}+\frac{4}{5}+\frac{5}{5}+\frac{6}{5}+\frac{7}{5}+\frac{8}{5}+\frac{9}{5}\)
\(=\frac{\left(9+1\right)\cdot9:2}{5}=\frac{45}{5}=9\)
\(A=11+14+17+...+62+65\)
Số số hạng của \(A\)là
\(\left(65-11\right)\div3+1=19\)(số hạng)
Tổng của \(A\)là:
\(\left(11+65\right)\times19\div2=722\)
Đáp số: 722
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+.....+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{9}-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\frac{10}{11}\times\frac{1}{2}\)
\(B=\frac{5}{11}\)
\(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{3}{154}\)
\(C=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(C=\left(\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+....+\frac{3}{11}-\frac{3}{14}\right)\div3\)
\(C=\left(\frac{3}{2}-\frac{3}{14}\right)\div3\)
\(C=\frac{9}{7}\div3\)
\(C=\frac{3}{7}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(B=1-\frac{1}{11}\)
\(B=\frac{10}{11}\)