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22 tháng 6 2018
\(1.\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\text{|}\sqrt{5}-\sqrt{2}\text{|}-\text{|}\sqrt{5}+\sqrt{2}\text{|}=-2\sqrt{2}\)\(2.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{|}2\sqrt{2}+\sqrt{5}\text{|}+\text{|}2\sqrt{2}-\sqrt{5}\text{|}=4\sqrt{2}\)\(3.\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{5}\right)\text{|}\sqrt{5}-\sqrt{2}\text{|}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
7 tháng 7 2018
\(1.\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\)
\(2.\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)
\(3.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}-\sqrt{3}\)
\(4.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(5.\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{4+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
\(6.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
5 tháng 8 2018
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
KT
5 tháng 8 2018
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
22 tháng 6 2018
\(1.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\text{ |}3\sqrt{2}-\sqrt{3\text{ }}\text{ |}=3\sqrt{2}-\sqrt{3}\)\(2.\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{9-2.3\sqrt{5}+5}+\sqrt{9+2.3\sqrt{5}+5}=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(3+\sqrt{5}\right)^2}=\text{ |}3-\sqrt{5}\text{ |}+\text{ |}3+\sqrt{5}\text{ |}=6\)\(3.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2.\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{ |}2\sqrt{2}+\sqrt{5}\text{ |}+\text{ |}2\sqrt{2}-\sqrt{5}\text{ |}=4\sqrt{2}\)\(4.\) Tương tự nhé bạn.
22 tháng 6 2018
Bn ơi câu 1 cái chỗ dấu bằng thứ 1 ák lm v đc ko
\(\sqrt{21-6\sqrt{6}}=\)\(2.\sqrt{18}\sqrt{3}\)
21 tháng 6 2018
\(a.\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{4+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\sqrt{\left(2+\sqrt{3}\right)^3}\) = \(\left(2-\sqrt{3}\right)\) | \(2+\sqrt{3}\) | = \(4-3=1\)
\(b.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\) \(=\) | \(2\sqrt{2}+\sqrt{5}\) | \(+\) | \(2\sqrt{2}-\sqrt{5}\) | \(=4\sqrt{2}+2\sqrt{5}\)
\(c.\sqrt{7-3\sqrt{5}}=\dfrac{\sqrt{14-2.3\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{9-2.3\sqrt{5}+5}}{\sqrt{2}}=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{2}}\)\(=\) \(\dfrac{\text{ |}3-\sqrt{5}\text{ |}}{\sqrt{2}}\) \(=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
\(d.\) Tương Tự nhé bạn.
6 tháng 12 2022
a: \(=2\cdot\sqrt{\dfrac{18-2\sqrt{77}}{4}}-\sqrt{20+6\sqrt{11}}\)
\(=\sqrt{11}-\sqrt{7}-\sqrt{11}-3=-\sqrt{7}-3\)
b: B=\(=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{7+\sqrt{13}}{18}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Đặt \(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)
\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
=>\(C=\sqrt{5}+1\)
\(B=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{14+2\sqrt{13}}{36}}+\sqrt{5}+1\)
\(=\dfrac{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}{6}+\sqrt{5}+1\)
=(13-1)/6+căn5+1
=3+căn5
7 tháng 10 2018
+) ta có : \(A=\sqrt{13+4\sqrt{10}}-\sqrt{13-4\sqrt{10}}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}-2\sqrt{2}+\sqrt{5}=2\sqrt{5}\) (sữa đề)
+) ta có : \(B=\sqrt{\dfrac{3-2\sqrt{2}}{17-12\sqrt{2}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(3-2\sqrt{2}\right)^2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{1}{\sqrt{2}-1}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{\dfrac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{2}+1+2-\sqrt{3}=3-\sqrt{3}+\sqrt{2}\) (sữa đề )
+) đk : \(x\ne-3\)
ta có : \(C=\dfrac{\sqrt{x^2+6x+9}}{x+3}=\dfrac{\sqrt{\left(x+3\right)^2}}{x+3}=\dfrac{\left|x+3\right|}{x+3}\)
\(\left[{}\begin{matrix}C=1\left(x>-3\right)\\C=-1\left(x< -3\right)\end{matrix}\right.\)
+) \(m\ge\dfrac{5}{2}\)
ta có : \(D=\sqrt{2m+4+6\sqrt{2m-5}}-\sqrt{2m-5}\)
\(=\sqrt{\left(\sqrt{2m-5}+3\right)^2}-\sqrt{2m-5}=\left|\sqrt{2m-5}+3\right|-\sqrt{2m-5}\)
\(\Leftrightarrow\left[{}\begin{matrix}C=3\left(m\ge7\right)\\C=-3-2\sqrt{2m-5}\left(\dfrac{5}{2}\le m\le7\right)\end{matrix}\right.\)