\(\sqrt{13}-\sqrt{12}và\sqrt{12}-\sqrt{11}\)
K
Khách
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8 tháng 8 2017
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
22 tháng 6 2017
C = \(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{12+2\sqrt{\left(\sqrt{13}+1\right)^2}}-\sqrt{\left(\sqrt{11}+1\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{14+2\sqrt{13}}-\left(\sqrt{11}+1\right)\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{13}+\sqrt{11}\right)\)
C \(\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)\) = \(13-11\) = \(2\)
TH
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}-\sqrt{12+2\sqrt{11}}}\right)\left(\sqrt{11}+\sqrt{3}\right)\)
3
5 tháng 10 2020
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)
B
1
10 tháng 10 2021
\(A=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
\(B=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)
mà \(\sqrt{12}+\sqrt{11}< \sqrt{14}+\sqrt{13}\)
nên A>B
NV
1
10 tháng 10 2017
ta xét hiệu A - B= \(\left(\sqrt{10}+\sqrt{13}\right)-\left(\sqrt{11}+\sqrt{12}\right)\) = \(\left(\sqrt{13}-\sqrt{12}\right)-\left(\sqrt{11}-\sqrt{10}\right)\)
\(\le\sqrt{13-12}-\sqrt{11-10}=1-1=0\)
6 tháng 7 2021
\(A=\sqrt{12+\sqrt{12+\sqrt{12}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{12+\sqrt{12+\sqrt{16}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}\)\(=7\)
\(B=\sqrt{14}+\sqrt{11}>\sqrt{13,69}+\sqrt{10,89}=7\)
\(\Rightarrow A< B\)
H
6 tháng 7 2021
Ta có:
\(12< 16\Rightarrow\sqrt{12}< \sqrt{16}=4\\ 6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)
\(\Rightarrow A< \sqrt{12+\sqrt{12+4}}+\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{12+4}+\sqrt{6+3}=4+3=7\) (1)
Lại có :
\(B=\sqrt{14}+\sqrt{11}\Rightarrow B^2=25+2\sqrt{14.11}=25+2\sqrt{154}>25+2\sqrt{144}=25+2.12=49=7^2\)
Mà B > 0
\(\Rightarrow B>7\) (2)
Từ (1),(2) suy ra A<B
14 tháng 3 2021
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{xy}+\sqrt{y}=11+12\sqrt{13}\\x+y=134\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)=12+12\sqrt{13}\\x+y=134\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}+1=a\\\sqrt{y}+1=b\end{matrix}\right.\) \(\left(a,b>0\right)\)
\(Hpt\Leftrightarrow\left\{{}\begin{matrix}ab=12+12\sqrt{13}\\a^2+b^2-2\left(a+b\right)+2=134\end{matrix}\right.\)
\(\Leftrightarrow a^2+b^2+2ab-2\left(a+b\right)+1=134+12+12\sqrt{13}-1\)
\(\Leftrightarrow\left(a+b\right)^2-2\left(a+b\right)+1=145+12\sqrt{13}\)
\(\Leftrightarrow\left(a+b-1\right)^2=145+12\sqrt{13}\)
\(\Leftrightarrow a+b=\sqrt{145+12\sqrt{13}}+1\)
\(Hpt\Leftrightarrow\left\{{}\begin{matrix}ab=12+12\sqrt{13}\\a+b=\sqrt{145+12\sqrt{13}}+1\end{matrix}\right.\)
Số xấu quá nên dừng tại đây :D
PK
2
9 tháng 9 2021
a: \(1< \sqrt{2}\)
nên \(2< \sqrt{2}+1\)
b: \(2\sqrt{31}=\sqrt{124}\)
\(10=\sqrt{100}\)
mà 124>100
nên \(2\sqrt{31}>10\)
c: \(-3\sqrt{11}=-\sqrt{99}\)
\(-\sqrt{12}=-\sqrt{12}\)
mà 99>12
nên \(-3\sqrt{11}< -\sqrt{12}\)
\(\sqrt{13}-\sqrt{12}=\dfrac{\left(\sqrt{13}-\sqrt{12}\right)\left(\sqrt{13}+\sqrt{12}\right)}{\sqrt{13}+\sqrt{12}}=\dfrac{13-12}{\sqrt{13}+\sqrt{12}}=\dfrac{1}{\sqrt{13}+\sqrt{12}}\)
\(\sqrt{12}-\sqrt{11}=\dfrac{\left(\sqrt{12}-\sqrt{11}\right)\left(\sqrt{12}+\sqrt{11}\right)}{\sqrt{12}+\sqrt{11}}=\dfrac{12-11}{\sqrt{12}+\sqrt{11}}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
Dễ dàng nhận thấy \(\sqrt{13}+\sqrt{12}>\sqrt{12}+\sqrt{11}>0\)
\(\Rightarrow\dfrac{1}{\sqrt{13}+\sqrt{12}}< \text{}\text{}\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
Vậy \(\sqrt{13}-\sqrt{12}< \sqrt{12}-\sqrt{11}\)
\(\sqrt{13}-\sqrt{12}và\sqrt{12}-\sqrt{11}\)
\(\sqrt{13}+\sqrt{11}và\sqrt{12}+\sqrt{12}\)
=> \(\left(\right)\sqrt{13}+\sqrt{11}\left(\right)^2và\left(\right)\sqrt{12}+\sqrt{12}\left(\right)^2\)
=>24+2\(\sqrt{13\cdot11}\) và 24+2*12
=2\(\sqrt{12^2-1}\) và 2*12
=>\(\sqrt{13}+\sqrt{11}< \sqrt{12}+\sqrt{12}\)
=> \(\sqrt{13}-\sqrt{12}< \sqrt{12}-\sqrt{11}\)