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\(\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-81^2\right)\\ =\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left[3^8-\left(3^4\right)^2\right]\\ =\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-3^8\right)\\ =\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right).0=0\)
\(\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-81^2\right)=\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right)\left(3^8-3^8\right)=\left(1^2+2^3+3^4+4^5\right)\left(1^3+2^3+3^3+4^3\right).0=0\)
Ta có: \(B=\dfrac{11}{13}\cdot\dfrac{40}{13}+\dfrac{37}{5}:\dfrac{7}{33}-\left(\dfrac{11}{13}+\dfrac{33}{7}:\dfrac{5}{2}\right)\)
\(=\dfrac{440}{169}+\dfrac{37}{5}\cdot\dfrac{33}{7}-\dfrac{11}{13}-\dfrac{33}{7}\cdot\dfrac{2}{5}\)
\(=\dfrac{297}{169}+\dfrac{33}{7}\cdot\left(\dfrac{37}{5}-\dfrac{2}{5}\right)\)
\(=\dfrac{297}{169}+\dfrac{33}{7}\cdot7\)
\(=\dfrac{297}{169}+33=\dfrac{5874}{169}\)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
\(A=\frac{7}{3\times13}+\frac{7}{13\times23}+...+\frac{7}{53\times63}\)
\(A=\frac{7}{10}.\left[\left(\frac{1}{3}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{23}\right)+....+\left(\frac{1}{53}-\frac{1}{63}\right)\right]\)
\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{53}-\frac{1}{63}\right)\)
\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(A=\frac{7}{10}.\frac{20}{63}\)
\(A=\frac{2}{9}\)
A=7*(1/3*13+1/13*23+1/23*33+1/33*43+1/43*53+1/53*63)
A=7/10(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43+1/43-1/53+1/53-1/63)
A=7/10*(1/3-1/63)
A=7/10*20/63
A=2/9
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-