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10 tháng 8 2020

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

21 tháng 10 2020

Ta có: \(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)

\(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}+\sqrt{2019}\right)\left(\sqrt{2020}-\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)

Do \(\frac{1}{\sqrt{2021}+\sqrt{2020}}< \frac{1}{\sqrt{2020}+\sqrt{2019}}\) => \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)

9 tháng 5 2021

ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)

Đặt \(\sqrt{x-2019}=a,......\)

Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)

- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)

\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)

- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )

Vậy ...

 

9 tháng 6 2019

a) Ta có: \(2\sqrt{5}=\sqrt{20}>\sqrt{7}\)

b) Ta có: \(4\sqrt{5}=\sqrt{80}< \sqrt{216}=6\sqrt{6}\)

\(\Rightarrow-4\sqrt{5}>-6\sqrt{6}\)

c) Ta có: \(\sqrt{2020}-\sqrt{2018}>0>\sqrt{2019}-\sqrt{2021}\)

14 tháng 7 2021

 \(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)

Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)

=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)

 \(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)

Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4

10 tháng 10 2020

Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)

\(=\frac{2020-2019}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)

\(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)

\(=\frac{2021-2020}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)

\(\sqrt{2020}+\sqrt{2019}< \sqrt{2021}+\sqrt{2020}\)

\(\Rightarrow\) \(\frac{1}{\sqrt{2020}+\sqrt{2019}}>\frac{1}{\sqrt{2021}+\sqrt{2020}}\)

Hay \(\sqrt{2020}-\sqrt{2019}>\sqrt{2021}-\sqrt{2020}\)

Chúc bn học tốt!