Cho \(A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}\)
và \(B=2\sqrt{2}+2\sqrt{4}+2.\sqrt{6}+....+2.\sqrt{18}+\sqrt{20}\)
So sánh A và B
Cho \(A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2.\sqrt{19}\)
và \(B=2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}\)
So sánh A và B
Cho A= \(2\sqrt{1}+2\sqrt{3}+...+2\sqrt{19}\) và B=\(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}\)
So sánh A và B
các bạn giải cho mình bài toán này với so sánh: A=\(2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+.......+2\sqrt{19}\)và B= \(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+....+2\sqrt{18}+\sqrt{20}\)
so sánh các số sau:
\(A=2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}\)
\(\text{Và }B=2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{20}\)
Xét hiệu :
\(A-B=2\left(\sqrt{1}-\sqrt{2}\right)+2.\left(\sqrt{3}-\sqrt{4}\right)+...+2\left(\sqrt{19}-\sqrt{20}\right)\)
Mà: \(\sqrt{1}
So sánh:
\(A=2\sqrt{1}+2\sqrt{3}+...+2\sqrt{19}\)
\(B=2\sqrt{2}+2\sqrt{4}+...+2\sqrt{18}+\sqrt{20}\)
So sánh các số sau:
A=\(2\sqrt{1}+2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}\)
B=\(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}\)
Ta có : \(\sqrt{3}-\sqrt{2}=\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1}{\sqrt{3}+\sqrt{2}}>\dfrac{1}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}\Rightarrow\sqrt{3}-\sqrt{2}>\sqrt{4}-\sqrt{3}\Rightarrow2\sqrt{3}>\sqrt{4}+\sqrt{2}\)
Làm tương tự : \(2\sqrt{5}>\sqrt{4}+\sqrt{6};2\sqrt{7}>\sqrt{6}+\sqrt{8},...,2\sqrt{19}>\sqrt{18}+\sqrt{20}\)
Cộng từng BĐT trên , ta được :
\(2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}>\sqrt{4}+\sqrt{2}+\sqrt{4}+\sqrt{6}+...+\sqrt{18}+\sqrt{20}=2\sqrt{4}+2\sqrt{6}+...+2\sqrt{18}+\sqrt{20}+\sqrt{2}\)
\(\Leftrightarrow A-2\sqrt{1}>B-\sqrt{2}\)
\(\Leftrightarrow A-B>2-\sqrt{2}>0\Rightarrow A>B\)
a) \(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{15-6\sqrt{6}}\)
b) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{19+2\sqrt{18}}\)
c) \(\sqrt{9+4\sqrt{5}}-\sqrt{\left(1-\sqrt{5}^2\right)}\)
\(a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3\)
a) \(=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=-1+\sqrt{6}\)
b) \(=\left|3-2\sqrt{2}\right|+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\)
c) \(=\sqrt{\left(\sqrt{5}+2\right)^2}-\left|1-\sqrt{5}\right|=\sqrt{5}+2+1-\sqrt{5}=3\)
a)\(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)=2\(\sqrt{6}-4+3-\sqrt{6}\)=\(\sqrt{6}-1\)
b)\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{18}\right)^2}\)=3-2\(\sqrt{2}+1+3\sqrt{2}\)=4+\(\sqrt{2}\)
c)\(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)=2+\(\sqrt{5}+1-\sqrt{5}\)=3
So sánh A=\(2\sqrt{1}\)+\(2\sqrt{3}+2\sqrt{5}+...+2\sqrt{19}+2\sqrt{21}với\)
B=\(\sqrt{2}+2\sqrt{4}+2\sqrt{6}+...+2\sqrt{20}+\sqrt{22}\)
So sánh A = 2\(\sqrt{1}+2\sqrt{3}+2\sqrt{5}+2\sqrt{7}+2\sqrt{9}+2\sqrt{11}+2\sqrt{13}+2\sqrt{15}+2\sqrt{17}+2\sqrt{19}\) và B = \(2\sqrt{2}+2\sqrt{4}+2\sqrt{6}+2\sqrt{8}+2\sqrt{10}+2\sqrt{12}+2\sqrt{14}+2\sqrt{16}+2\sqrt{18}+2\sqrt{20}\)