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\(M=\frac{10^{2018}+1}{10^{2019}+1}\)
\(\Rightarrow10M=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)
\(N=\frac{10^{2019}+1}{10^{2020}+1}\)
\(\Rightarrow10N=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)
Ta co: \(\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}\) ma \(1=1\)
\(\Rightarrow1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}\)
\(\Rightarrow10M>10N\)
\(\Rightarrow M>N\)
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)
\(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)
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\(1=1\)
\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
đặt 22018 = a ; 32019 = b ; 52020 = c
Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
\(\Rightarrow A>1>\frac{3}{4}>B\)
Mình chỉ biết cách tính B thôi, đây nhé:
B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)
\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2018}{5^{2018}}\)
\(5A=1+\frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+...+\frac{2018}{5^{2017}}\)
Trừ dưới cho trên:
\(4A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2017}}-\frac{2018}{5^{2018}}\)
Đặt \(C=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2017}}\) (1)
\(\Rightarrow4A=C-\frac{2018}{5^{2018}}< C\Rightarrow A< \frac{1}{4}C\)
Ta có: \(\frac{1}{5}C=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2017}}+\frac{1}{5^{2018}}\) (2)
Trừ (1) cho (2):
\(\frac{4}{5}C=1-\frac{1}{5^{2018}}\Rightarrow C=\frac{5}{4}-\frac{1}{4.5^{2017}}< \frac{5}{4}\)
\(\Rightarrow A< \frac{1}{4}C< \frac{1}{4}.\frac{5}{4}=\frac{5}{16}< \frac{2018}{2019}\)
\(\Rightarrow A< B\)
\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)
\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)
\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)
\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)
Nhận thấy : \(1+\frac{2017}{2018^{2019}+1}>1-\frac{4034}{2018^{2020}-2}\Leftrightarrow2M>2N\Leftrightarrow M>N\)
Từ đề bài, ta suy ra:
So sánh hai biểu thức
\(M=\left(2018^{2018}+1\right)\cdot\left(2018^{2020}-2\right)\)(1)
\(N=\left(2018^{2019}-2\right)\cdot\left(2018^{2019}+1\right)\)(2)
Xét biểu thức M và N, ta suy ra:
\(M=\left(2018^{2019}-2017\right)\cdot\left(2019^{2019}+2016\right)\)
\(N=\left(2018^{2019}-2017\right)\cdot\left(2018^{2018}-2016\right)\)
Nhận thấy (20192019+2016)>(20182018-2016) nên M>N
Vậy M>N.
P/s:Mình đây không phải top 10 tuần nên bài có thể sai sót, mong bạn tham khảo:)))