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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 \(\frac{1}{\sqrt{2021}+\sqrt{2020}}< \frac{1}{\sqrt{2020}+\sqrt{2019}}\) => \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)
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
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!
TA XÉT PHÂN THỨC TỔNG QUÁT SAU:
\(A=\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}\)
\(A=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(A=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)
\(A=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
\(A=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
THAY LẦN LƯỢT CÁC GIÁ TRỊ n từ 1 => 2021 vào ta được:
=> \(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
=> \(A=1-\frac{1}{\sqrt{2021}}=\frac{\sqrt{2021}-1}{\sqrt{2021}}\)
VẬY \(A=\frac{\sqrt{2021}-1}{\sqrt{2021}}.\)
Ta có: \(\frac{1}{\left(a-1\right)\sqrt{a}+a.\sqrt{a-1}}=\frac{a-\left(a-1\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}\)
\(=\frac{\left(\sqrt{a}-\sqrt{a-1}\right)\left(\sqrt{a}+\sqrt{a-1}\right)}{\sqrt{a}.\sqrt{a-1}.\left(\sqrt{a}+\sqrt{a-1}\right)}=\frac{\sqrt{a}-\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}\)
\(=\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a-1}}-\frac{\sqrt{a-1}}{\sqrt{a}.\sqrt{a-1}}=\frac{1}{\sqrt{a-1}}-\frac{1}{\sqrt{a}}\)
Thay lần lượt các giá trị của a bằng \(2;3;4;........;2021\)ta được:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.........+\frac{1}{\sqrt{2020}}-\frac{1}{\sqrt{2021}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2021}}=1-\frac{1}{\sqrt{2021}}\)
Đk: \(\forall x\in R\)
Ta có:\(\sqrt{x^2+1-2x}+\sqrt{x^2+4x+4}=\sqrt{1+2020^2+\frac{2020^2}{2021^2}}+\frac{2020}{2021}\)
<=> \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=\sqrt{1+2020^2+2.2020+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(1+2020\right)^2+\frac{2020^2}{2021^2}-2.2020}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=\frac{2021^2-2020}{2021}+\frac{2020}{2021}\)
<=> \(\left|x-1\right|+\left|x+2\right|=2021\)
Lập bảng xét dầu
x -2 1
x - 1 - | - 0 +
x + 2 - 0 + | -
Xét các TH xảy ra :
TH1: x \(\le\)-2 => pt trở thành: 1 - x - x - 2 = 2021
<=> -2x = 2022 <=> x = -1011 (tm)
TH2: \(-2< x\le1\) => pt trở thành: 1 - x + x + 2 = 2021
<=> 0x = 2018 (vô lí) => pt vô nghiệm
TH3: \(x>1\) => pt trở thành: x - 1 + x + 2 = 2021
<=> 2x = 2020 <=> x = 1010 (tm)
Vậy S = {-1011; 1010}
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: \(\sqrt{2020}-\sqrt{2018}>0>\sqrt{2019}-\sqrt{2021}\)