18 : 3 (1+2) =?
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\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
\(A=\left(\frac{2.18-1}{18}\right)\left(\frac{2.18-2}{18}\right)\left(\frac{2.18-3}{18}\right)....\left(\frac{2.18-35}{18}\right)\left(\frac{2.18-36}{18}\right)\left(\frac{2.18-37}{18}\right)...\left(\frac{2.18-100}{18}\right)\)
\(=\frac{35}{18}.\frac{34}{18}.\frac{33}{18}...\frac{1}{18}.\frac{0}{18}.\frac{-1}{18}...\frac{-64}{18}=0\)
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
Lời giải:
Đặt \(A=1+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{18}{3^{18}}\)
\(\Rightarrow 3A=3+1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{18}{3^{17}}\)
\(\Rightarrow 3A-A=3+\frac{2-1}{3}+\frac{3-2}{3^2}+\frac{4-3}{3^3}+..+\frac{18-17}{3^{17}}-\frac{18}{3^{18}}\)
\(2A=3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{17}}-\frac{18}{3^{18}}\)
\(\Rightarrow 6A=9+1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{16}}-\frac{18}{3^{17}}\)
\(\Rightarrow 6A-2A=7-\frac{18}{3^{17}}-\frac{1}{3^{17}}+\frac{18}{3^{18}}\)
\(\Leftrightarrow 4A=7+\frac{18}{3^{18}}-\frac{19}{3^{17}}=7-\frac{39}{3^{18}}\)
\(\Rightarrow A=\frac{1}{4}\left(7-\frac{39}{3^{18}}\right)\)
\(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\) = \(\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+1\)
= \(\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{20}\)
=\(20.\left(\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}+\frac{1}{20}\right)\)
=\(20.\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}+...+\frac{1}{2}\right)\)
Vì tử số gấp 20 lần mẫu số nên phân số này bằng 20
2 nha bn
bằng 2 bn ạ