\(\frac{2}{3}+\frac{20}{3}=???\)
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Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
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\(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+......+\frac{2+3+4+...+20}{1+2+3+4+...+20}\)
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\(S=3+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^9}\)=> \(2S=6+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\)
=>\(2S-S=\left(6+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^9}\right)\)
=>\(S=3+\frac{3}{2}-\frac{3}{2^9}\) tự tính nốt
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\(x+\frac{3+\frac{3}{20}+\frac{3}{13}+\frac{3}{2013}}{5+\frac{5}{20}+\frac{5}{13}+\frac{5}{2013}}=\frac{9}{2}\)
\(x+\frac{3\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}{5\times\left(1+\frac{1}{20}+\frac{1}{13}+\frac{1}{2013}\right)}=\frac{9}{2}\)
\(x+\frac{3}{5}=\frac{9}{2}\)
\(x=\frac{39}{10}\)
\(x=3,9\)
Bài làm
\(\frac{2}{3}+\frac{20}{3}=\frac{2+20}{3}=\frac{22}{3}\)
\(\frac{2}{3}+\frac{20}{3}=\frac{2+20}{3}\)
\(=\frac{22}{3}\)
hok tốt