Tính B
B=\(\frac{1}{2019}+\frac{2}{2019}+\frac{3}{2019}+\frac{4}{2019}+...+\frac{2019}{2019}\)
Những câu hỏi liên quan
So sánh \(A=\frac{2^{2019}}{2^{2019}+3^{2019}}+\frac{3^{2019}}{3^{2019}+5^{2019}}+\frac{5^{2019}}{5^{2019}+2^{2019}}\)với \(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
Tham khảo
https://hoc24.vn/hoi-dap/question/814814.html
B=11.2+13.4+15.6+....+12019.2020
⇒2B=21.2+23.4+25.6+....+22019.2020
<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020
2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020
2B<1+12−13+13−14+...+12019−12020
2B<1+12−12020<1+12
B<34
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Đặt 22018=a;32019=b;52020=c(a,b,c>0)
A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1
⇒A>1>34>B
So sánh hai số A và B biết :
A = \(-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}\)
B = \(-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\)
Help me , pleaseeeeeeeeee
\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)
=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)
\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)
=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)
=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)
=>\(A>B\)
cách này mình tự nghĩ
thank you \(v\text{er}y^{1000000000000}\)much
tính giá trị biểu thức\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2019}{1}+\frac{2019}{2}+\frac{2017}{3}+...+\frac{1}{2019}}\)
Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
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So sánh A và B biết :
A = \(-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}\)
B = \(-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\)
Giúp mk nha mn
Lời giải:
\(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
Dễ thấy $0< 2019^2< 2019^4\Rightarrow \frac{4}{2019^2}> \frac{4}{2019^4}$
$\Rightarrow A-B=\frac{4}{2019^2}-\frac{4}{2019^4}>0$
$\Rightarrow A>B$
Rút gọn biểu thức S = \(\frac{2019}{2\sqrt{1}+1\sqrt{2}}+\frac{2019}{3\sqrt{2}+2\sqrt{3}}+\frac{2019}{4\sqrt{3}+3\sqrt{4}}+...+\frac{2019}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Mk chỉ cần kết quả thôi , cảm ơn nhiều ạ
Tính tổng :
\(\frac{2019}{210}+\frac{2019}{280}+\frac{2019}{360}+\frac{2019}{450}+\frac{2019}{550}\)
\(\frac{2019}{210}+\frac{2019}{280}+\frac{2019}{360}+\frac{2019}{450}+\frac{2019}{550}\)
\(=\frac{673}{70}+\frac{2019}{280}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{70}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{2692}{280}+\frac{2019}{280}\right]+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\frac{673}{40}+\frac{673}{120}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{40}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{2019}{120}+\frac{673}{120}\right]+\frac{673}{150}+\frac{2019}{550}\)
\(=\frac{673}{30}+\frac{673}{150}+\frac{2019}{550}\)
\(=\left[\frac{673}{30}+\frac{673}{150}\right]+\frac{2019}{550}\)
\(=\frac{673}{25}+\frac{2019}{550}=\frac{14806}{550}+\frac{2019}{550}=\frac{16825}{550}=\frac{673}{22}\)
P/S : Các a chị check dùm em ạ
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cho các số a,b,c khác 0 sao cho \(a+b=c+\frac{1}{2019}\)và \(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+2019\)
tính giá trị của \(P=\left(a^{2019}+b^{2019}-c^{2019}\right)\left(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}-\frac{1}{c^{2019}}\right)\)
\(a+b=c+\frac{1}{2019}\Leftrightarrow a+b-c=\frac{1}{2019}\Leftrightarrow\frac{1}{a+b-c}=2019\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+2019\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=2019\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=\frac{1}{a+b-c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b-c}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{a+b}{c\left(a+b-c\right)}\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)=\left(a+b\right)ab\)
\(\Leftrightarrow c\left(a+b-c\right)\left(a+b\right)-ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ca+bc-c^2-ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a-c\right)-b\left(a-c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(c-b\right)\left(a-c\right)=0\)
=>a=-b hoặc c=b hoặc a=c
không mất tính tổng quát, giả sử a=-b, ta có:
\(P=\left(-b^{2019}+b^{2019}-c^{2019}\right)\left(-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}-\frac{1}{c^{2019}}\right)=\left(-c\right)^{2019}\cdot\left(\frac{-1}{c}\right)^{2019}=1\)
tương tư với các trường hợp khác ta cũng có P=1
Vậy P=1
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Chứng minh nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
thì \(\frac{1}{x^{2019}}+\frac{1}{y^{2019}}+\frac{1}{z^{2019}}=\frac{1}{x^{2019}+y^{2019}+z^{2019}}\)
Xem chi tiết
Cho A= \(\frac{2020}{2019^2+1}+\frac{2020}{2019^2+2}+\frac{2020}{2019^2+3}+...+\frac{2020}{2019^2+2019}\)
Chứng minh rằng A ko thể là số tự nhiên.
Ta có bài toán tổng quát sau:Chứng minh rằng tổng \(A=\frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+....+\frac{n+1}{n^2+n}\)(n số hạng và n>1) không phải là số nguyên dương ta có:
\(1=\frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+...+\frac{n+1}{n^2+3}< \frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+....+\frac{n+1}{n^2+n}< \frac{n+1}{n^2}+\frac{n+1}{n^2}\)\(+....+\frac{n+1}{n^2}=2\)
Do đó A không phải là số nguyên dương với n=2019 thì ta có bài toán đã cho