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* 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
Mk lm đc câu a thôi nhé !
A= 150-(100-99+98-97+...-3+2-1)
từ 1-100 có 100 SH. Ta nhóm 4 số vs nhau như sau : (100-00+98-87)+(...)+(4-3+2-1)
Có tất cả số nhóm là : 100:4=25 nhóm. Mà mỗi nhóm ta tính có kết quả là 2, vậy tao có
A=150-(2.25)
A=150-50
A=100
S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{18^2}\)+\(\frac{1}{19^2}\)
S<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{17.18}\)+\(\frac{1}{18.19}\)
S<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{17}\)-\(\frac{1}{18}\)+\(\frac{1}{18}\)-\(\frac{1}{19}\)
S<1-\(\frac{1}{19}\)
\(\Rightarrow\)S<\(\frac{18}{18}\)
Lời giải:
Đặt \(A=1+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{18}{3^{18}}\)
\(\Rightarrow 3A=3+1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{18}{3^{17}}\)
\(\Rightarrow 3A-A=3+\frac{2-1}{3}+\frac{3-2}{3^2}+\frac{4-3}{3^3}+..+\frac{18-17}{3^{17}}-\frac{18}{3^{18}}\)
\(2A=3+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{17}}-\frac{18}{3^{18}}\)
\(\Rightarrow 6A=9+1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{16}}-\frac{18}{3^{17}}\)
\(\Rightarrow 6A-2A=7-\frac{18}{3^{17}}-\frac{1}{3^{17}}+\frac{18}{3^{18}}\)
\(\Leftrightarrow 4A=7+\frac{18}{3^{18}}-\frac{19}{3^{17}}=7-\frac{39}{3^{18}}\)
\(\Rightarrow A=\frac{1}{4}\left(7-\frac{39}{3^{18}}\right)\)
\(A=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^2\left(5.2^{28}.3^{18}\right)-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^2-1}{1-7}\)
\(A=\frac{3}{-6}=\frac{1}{-2}=-\frac{1}{2}\)
5^x.5^(x+1).5^(x+2).2^18<=10^18=2^18.5^18
x+x+1+x+2<=18
3x+3<=18
x+1<=6
x<=5
nếu tự nhiện thì x=(0,1,2,3,4,5)
\(A=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^{14}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.\left(2^2\right)^9.\left(3^2\right)^9.\left(2^2\right)^6-2.\left(2^2\right)^{14}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{18}}{5.2^{28}.3^{18}-7.2.2^{28}.3^{18}}\)
\(A=5.\frac{\left(2^{18}.2^{12}\right).3^{18}.2^{29}.3^{18}}{2^{28}.3^{18}.\left(5-7-2\right)}\)
\(A=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{5.2.2^{29}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2^{20}.2^{18}.\left(5.2-1\right)}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.\left(-9\right)}\)
\(A=\frac{2.11}{11.\left(-2\right)}\)
\(A=-2\)
Vậy : A = -2
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@Nguyễn Đức Huy, bạn ấn vào đọc tiếp thì xem đc nhé!