13/7 + 33/38
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\(\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-81^2\right)\\ =\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left[3^8-\left(3^4\right)^2\right]\\ =\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-3^8\right)\\ =\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right).0=0\)
\(\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-81^2\right)=\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-3^8\right)=\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right).0=0\)
1/
$M=\frac{1}{3.8}+\frac{1}{8.13}+\frac{1}{13.18}+....+\frac{1}{33.38}$
$5M=\frac{8-3}{3.8}+\frac{13-8}{8.13}+\frac{18-13}{13.18}+...+\frac{38-33}{33.38}$
$=\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+....+\frac{1}{33}-\frac{1}{38}$
$=\frac{1}{3}-\frac{1}{38}=\frac{35}{114}$
$\Rightarrow M=\frac{35}{114}:5=\frac{7}{114}$
2/
$N=\frac{1}{3.10}+\frac{1}{10.17}+\frac{1}{17.24}+\frac{1}{24.31}+\frac{1}{31.38}$
$7N=\frac{10-3}{3.10}+\frac{17-10}{10.17}+\frac{24-17}{17.24}+\frac{31-24}{24.31}+\frac{38-31}{31.38}$
$=\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+\frac{1}{17}-\frac{1}{24}+\frac{1}{24}-\frac{1}{31}+\frac{1}{31}-\frac{1}{38}$
$=\frac{1}{3}-\frac{1}{38}=\frac{35}{114}$
$\Rightarrow N=\frac{35}{114}:7=\frac{5}{114}$
Hướng dẫn giải:
a) 93 + 39 + 7 = (93 + 7) + 39 = 100 + 39 = 139.
89 + 85 + 11 = (89 + 11) + 85 = 100 + 85 = 185
184 + 66 + 16 = (184 + 16) + 66 = 200 + 66 = 266
b) 23 + 52 + 47 + 98 = (23 + 47) + (52 + 98) = 70 + 150 = 220
16 + 490 + 84 + 10 = (16 + 84) + (490 + 10) = 100 + 500 = 600
\(\dfrac{33}{38}:\dfrac{11}{19}=\dfrac{33\times19}{38\times11}=\dfrac{3\times1}{2\times1}=\dfrac{3}{2}\)
Ta có: \(B=\dfrac{11}{13}\cdot\dfrac{40}{13}+\dfrac{37}{5}:\dfrac{7}{33}-\left(\dfrac{11}{13}+\dfrac{33}{7}:\dfrac{5}{2}\right)\)
\(=\dfrac{440}{169}+\dfrac{37}{5}\cdot\dfrac{33}{7}-\dfrac{11}{13}-\dfrac{33}{7}\cdot\dfrac{2}{5}\)
\(=\dfrac{297}{169}+\dfrac{33}{7}\cdot\left(\dfrac{37}{5}-\dfrac{2}{5}\right)\)
\(=\dfrac{297}{169}+\dfrac{33}{7}\cdot7\)
\(=\dfrac{297}{169}+33=\dfrac{5874}{169}\)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
TL
13/7+33/38=725/266
nha bn
HT
\(\frac{13}{7}+\frac{33}{38}=\frac{494}{266}+\frac{231}{266}=\frac{725}{266}\)
-HT-