so sanh
2019 x 2021 va 2020 x 2020
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\(\sqrt{2021}-\sqrt{2020}=\dfrac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\dfrac{1}{\sqrt{2021}+\sqrt{2020}}\) là nghịch đảo của \(\sqrt{2021}+\sqrt{2020}\) (đpcm)
\(\sqrt{2021}-\sqrt{2020}=\dfrac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)
\(=\dfrac{1}{\sqrt{2021}+\sqrt{2020}}\)(đpcm)
Đặt \(2020-x=u;x-2021=v\)thì \(u+v=-1\)
Phương trình trở thành \(\frac{u^2+uv+v^2}{u^2-uv+v^2}=\frac{19}{49}\Leftrightarrow30u^2+30v^2+68uv=0\)
\(\Leftrightarrow15\left(u+v\right)^2+4uv=0\Leftrightarrow4uv=-15\Leftrightarrow uv=\frac{-15}{4}\)
hay \(\left(2020-x\right)\left(x-2021\right)=-\frac{15}{4}\Leftrightarrow x^2-4041x+4082416,25=0\)
Dùng công thức nghiệm tìm được x = 2022, 5 hoặc x = 2018, 5
Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)
\(=\frac{2020-2019}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)
\(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)
\(=\frac{2021-2020}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
Vì \(\sqrt{2020}+\sqrt{2019}< \sqrt{2021}+\sqrt{2020}\)
\(\Rightarrow\) \(\frac{1}{\sqrt{2020}+\sqrt{2019}}>\frac{1}{\sqrt{2021}+\sqrt{2020}}\)
Hay \(\sqrt{2020}-\sqrt{2019}>\sqrt{2021}-\sqrt{2020}\)
Chúc bn học tốt!
\(2019\times2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2=2020\times2020\)
Ta có:
2019.2021=2019.(2020+1)=2019.2020+2019 (1)
Lại có:
2020.2020=(2019+1).2020=2019.2020+2020 (2)
Vì 2019.2020=2019.2020 mà 2019<2020
=>(1)<(2)
=>.....
Bài giải
Ta có : \(2019\text{ x }2021=2019\text{ x }2020+2019\)
\(2020\text{ x }2020=2019\text{ x }2020+2020\)
\(\text{Vì }2019\text{ x }2020+2019< 2019\text{ x }2020+2020\text{ }\Rightarrow\text{ }2019\text{ x }2021< 2020\text{ x }2020\)