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1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
a) \(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
b) \(\sqrt{13-4\sqrt{3}}=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+1}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}=2\sqrt{3}-1\)
c) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
d) \(\sqrt{3+2\sqrt{2}+\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{3+2\sqrt{2}+\sqrt{\left(2-\sqrt{2}\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}+2-\sqrt{2}}\)
\(=\sqrt{5+\sqrt{2}}\)
e) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
\(=2+\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)
\(=2+\sqrt{9-4\sqrt{5}}\)
\(=2+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2+\sqrt{5}-2=\sqrt{5}\)
f) đề sai nhé:
\(\sqrt{3a}.\sqrt{12a}=\sqrt{36a^2}=6a\)\(\left(a\ge0\right)\)
g) \(\sqrt{16a^2b^8}=4b^4\left|a\right|\)
h) \(\sqrt{7a}.\sqrt{63a^3}=\sqrt{441.a^4}=21a^2\)
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
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
\(1.\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\)
\(2.\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)
\(3.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}-\sqrt{3}\)
\(4.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(5.\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{4+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
\(6.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
a) \(\sqrt{2017}-2\sqrt{2016}=\sqrt{2017}-\sqrt{8064}< 0< \sqrt{2016}\)
b) \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=8=\sqrt{64}>\sqrt{61}\)
c) \(\left(\sqrt{2016}+\sqrt{2014}\right)^2=4030+\sqrt{2014.2016}\)
\(\left(2\sqrt{2015}^2\right)=4030+\sqrt{2015.2015}\)
C/m được: \(\sqrt{2014.2016}< \sqrt{2015.2015}\)
\(\Rightarrow\left(\sqrt{2016}+\sqrt{2014}\right)^2< \left(2\sqrt{2015}\right)^2\)
\(\Rightarrow\sqrt{2014}+\sqrt{2016}< 2\sqrt{2015}\)
d) \(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=7=8-1=\sqrt{64}-1< \sqrt{65}-1\)