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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 \(x=2020\)
\(A=2019\times2021=\left(2021-1\right)\times\left(2021+1\right)=2021^2-1< 2021^2=B.\)
\(\left(2x+1\right)^3=-125\)
\(< =>\left(2x+1\right)^3=\left(-5\right)^3\)
\(< =>2x+1=-5\)
\(< =>2x=-5-1=-6\)
\(< =>x=-\frac{6}{2}=-3\)
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
\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)
\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)
\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)
\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)
\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)
(x+2019)(x-2020)=0.
=> x+2019=0 hoặc x-2020=0.
+, x+2019=0. +, x-2020=0
x= 0-2019 x = 0+2020
x = -2019. x = 2020.
Vậy: x thuộc{ -2019 ; 2020 }.
#Học tốt.
\(\Leftrightarrow\orbr{\begin{cases}x+2019=0\\x-2020=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2019\\x=2020\end{cases}}}\)