Ta có công thức A.B=0 suy ra A=0,B=0

Suy ra X-2019=0  ⟹X=0+2019 ⟹X=2019

X-2020=0 ⟹X=0+2020 ⟹X=2020

\(\left(x-2019\right).\left(x-2020\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2019=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2019\\x=2020\end{cases}}\)

Vậy \(x=2019\)hoặc \(x=2020\)

\(A=2019\times2021=\left(2021-1\right)\times\left(2021+1\right)=2021^2-1< 2021^2=B.\)

sai mất rồi nhưng dù sao cũng cảm ơn bn nhé

\(\left(x-2019\right)\left(x-2020\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2019=0\\x-2020=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2019\\x=2020\end{cases}}\)

Vậy \(x\in\left\{2019;2020\right\}\)

\(\left(2x+1\right)^3=-125\)

\(< =>\left(2x+1\right)^3=\left(-5\right)^3\)

\(< =>2x+1=-5\)

\(< =>2x=-5-1=-6\)

\(< =>x=-\frac{6}{2}=-3\)

Bài làm:

a) \(\left(2x+1\right)^3=-125\)

\(\Leftrightarrow\left(2x+1\right)^3=\left(-5\right)^3\)

\(\Rightarrow2x+1=-5\)

\(\Leftrightarrow2x=-6\)

\(\Rightarrow x=-3\)

b) \(\left(7-x\right)^2-\left(-11\right)=15\)

\(\Leftrightarrow\left(7-x\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}7-x=2\\7-x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=9\end{cases}}\)

c) \(2020^x-2019=-2019\)

\(\Leftrightarrow2020^x=0\)

=> ko tồn tại x thỏa mãn PT

Ta có: \(P=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(3P=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}\)

=>3P+P=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>4A=-1\(-\frac{1}{3^{2019}}\)

=>\(4A=\frac{-3^{2019}-1}{3^{2019}}\)

=>\(A=\frac{-3^{2019}-1}{4\cdot3^{2019}}\)

Ta có: \(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

\(=1+\frac{-3^{2019}-1}{4\cdot3^{2019}}-\frac{2020}{3^{2020}}=1+\frac{-3^{2020}-3-8080}{4\cdot3^{2020}}=1-\frac14-\frac{8083}{4\cdot3^{2020}}<\frac34\)

=>\(P<\frac{3}{16}\) (ĐPCM)

\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)

\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)

\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)