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Đặt \(A=\left(\sqrt{2018}+\sqrt{2020}\right)\)
\(\Rightarrow A^2=2018+2\sqrt{2018.2020}+2020=4038+\sqrt{4.2018.2020}=4038+\sqrt{4.\left(2019^2-1\right)}\)
Đặt \(B=2\sqrt{2019}=\sqrt{4.2019}\)
\(B^2=4.2019=2.2019+2.2019=4038+\sqrt{4.2019^2}\)
=> \(\sqrt{4.2019^2}>\sqrt{4.\left(2019^2-1\right)}\)
\(\Rightarrow A>B\Leftrightarrow\sqrt{2018}+\sqrt{2020}>2\sqrt{2019}\)
1,Ta có : \(\sqrt{11}-\sqrt{10}=\frac{11-10}{\sqrt{11}+\sqrt{10}}=\frac{1}{\sqrt{11}+\sqrt{10}}\)
\(\sqrt{6}-\sqrt{5}=\frac{6-5}{\sqrt{6}-\sqrt{5}}=\frac{1}{\sqrt{6}-\sqrt{5}}\)
Dễ thấy : \(11+10>6+5\Rightarrow\sqrt{11}+\sqrt{10}>\sqrt{6}+\sqrt{5}\)
từ đó suy ra : \(\frac{1}{\sqrt{11}+\sqrt{10}}< \frac{1}{\sqrt{6}+\sqrt{5}}\)( theo so sánh phân số có cùng tử )
Vậy...
2,\(\sqrt{2019}+\sqrt{2021}và2\sqrt{2020}\)
Giả sử : \(\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)
\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\) ( bình phương 2 vế )
\(\Leftrightarrow2019+2021+2\sqrt{2019.2021}< 4.2020\)
\(\Leftrightarrow4040+2\sqrt{2020^2-1^2}< 8080\)
\(\Leftrightarrow\)\(4040+\left(-4040\right)+2\left|2020-1\right|< 8080+\left(-4040\right)\)
( cộng cả hai vế với -4040)
\(\Leftrightarrow2.2019< 4040\)
\(\Leftrightarrow\frac{1}{2}.2.2019< 4040.\frac{1}{2}\)( nhân hai vế với 1/2)
\(\Leftrightarrow2019< 2020\) ( luôn đúng )
=> điều giả sử đúng
Vậy....
4,Ta có : \(\sqrt{2020}-\sqrt{2019}=\frac{2020-2019}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)
\(\sqrt{2019}-\sqrt{2018}=\frac{2019-2018}{\sqrt{2019}+\sqrt{2018}}=\frac{1}{\sqrt{2019}+\sqrt{2018}}\)
dễ thấy \(2020+2019>2019+2018\Rightarrow\sqrt{2020}+\sqrt{2019}>\sqrt{2019}+\sqrt{2018}\) Từ đó suy ra : \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2020}-\sqrt{2019}}\)
theo ss phân số có cùng tử
Vậy....
phần 5 làm tương tự như phần 4 nhé
a) Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)
\(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)
Vậy x < y
Ta có \(4m-m^2-5=-\left(m-2\right)^2-1< 0\) \(\forall m\)
\(\Rightarrow f\left(x\right)\) nghịch biến trên R \(\Rightarrow f\left(a\right)>f\left(b\right)\Leftrightarrow a< b\)
Mà \(2-\sqrt{2019}>2-\sqrt{2020}\Rightarrow f\left(2-\sqrt{2019}\right)< f\left(2-\sqrt{2020}\right)\)
Ta có: \(\sqrt{2018}-\sqrt{2019}\)
\(=\frac{\left(\sqrt{2018}-\sqrt{2019}\right)\left(\sqrt{2018}+\sqrt{2019}\right)}{\sqrt{2018}+\sqrt{2019}}\)
\(=\frac{2018-2019}{\sqrt{2018}+\sqrt{2019}}=\frac{-1}{\sqrt{2018}+\sqrt{2019}}\)
Ta có: \(\sqrt{2019}-\sqrt{2020}\)
\(=\frac{\left(\sqrt{2019}-\sqrt{2020}\right)\left(\sqrt{2019}+\sqrt{2020}\right)}{\sqrt{2019}+\sqrt{2020}}\)
\(=\frac{-1}{\sqrt{2019}+\sqrt{2020}}\)
Ta có: \(\sqrt{2018}+\sqrt{2019}< \sqrt{2019}+\sqrt{2020}\)
\(\Leftrightarrow\frac{1}{\sqrt{2018}+\sqrt{2019}}>\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
\(\Leftrightarrow\frac{-1}{\sqrt{2018}+\sqrt{2019}}< \frac{-1}{\sqrt{2019}+\sqrt{2020}}\)
hay \(\sqrt{2018}-\sqrt{2019}< \sqrt{2019}-\sqrt{2020}\)
Ta có: \(\left(\sqrt{2018}+\sqrt{2020}\right)^2=2018+2020+2\sqrt{2018.2020}\)
\(=4038+2\sqrt{\left(2019-1\right)\left(2019+1\right)}< 4038+2\sqrt{2019^2}\)
\(=4038+4038=8076\) (1)
Ta cũng có: \(\left(2\sqrt{2019}\right)^2=4.2019=8076\) (2)
Từ (1) và (2) \(\Rightarrow\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)
Xét \(\left(\sqrt{2018}+\sqrt{2020}\right)^2=2018+2020+2\sqrt{2018.2020}\)
\(=2019+2019+2\sqrt{\left(2019+1\right)\left(2019-1\right)}\)
\(=2.2019+2\sqrt{2019^2-1}\)
Có \(\sqrt{2019^2-1}< \sqrt{2019^2}\Rightarrow2\sqrt{2019^2-1}< 2.2019\)
\(\Rightarrow2.2019+2\sqrt{2019^2-1}< 2.2019+2.2019=4.2019=\left(2\sqrt{2019}\right)^2\)
\(\Rightarrow\left(\sqrt{2018}+\sqrt{2020}\right)^2< \left(2\sqrt{2019}\right)^2\)
\(\Leftrightarrow\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)