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a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
Thật ra tui cũng không rõ lắm đâu. Cậu thử nhân A với \(\dfrac{2019}{2020}\)rồi lại cộng lại với A thử coi nào <Chú Ý : chưa chắc đã đúng >
Ta có: \(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}=\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\)
\(\Rightarrow\left(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}\right)-\left(\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\right)=0\)
\(\Rightarrow\dfrac{x+1}{2017}+\dfrac{x+1}{2018}-\dfrac{x+1}{2019}-\dfrac{x+1}{2020}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}\right)=0\)
Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}>\dfrac{1}{2019}>\dfrac{1}{2020}>0\) nên
\(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}>0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)
=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)
\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)
=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)
=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)
=>\(A>B\)
cách này mình tự nghĩ
Tính [G(x) - f(x) ] = ( \(1-x^2+.....+x^{2020}\)) - (\(x^{2020}-x^{2019}+....-x+1\))
= (\(x^{2020}-x^{2019}+....-x+1\)) - (\(x^{2020}-x^{2019}+....-x+1\))
= 0
=> h(x) = [G(x) - f(x) ] * [G(x) + f(x) ]
= 0 * [G(x) + f(x) ]
= 0
\(A=\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)
\(B=\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
Ta có :
\(A-B=\dfrac{2020^{2018}-1}{2020^{2019}+2019}-\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
\(\Rightarrow A-B=\dfrac{\left(2020^{2018}-1\right)\left(2020^{2020}+2019\right)-\left(2020^{2019}+2019\right)\left(2020^{2019}+1\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{2020^{4038}+2019.2020^{2018}-2020^{2020}-2019-2020^{4038}-2020^{2019}-2019.2020^{2018}-2029}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{-\left(2020^{2020}+2020^{2019}+2.2019\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
mà \(\left\{{}\begin{matrix}-\left(2020^{2020}+2020^{2019}+2.2019\right)< 0\\\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)>0\end{matrix}\right.\)
\(\Rightarrow A-B< 0\)
\(\Rightarrow A< B\)
Vậy ta được \(A< B\)