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24 tháng 9 2016
Cho mình sửa đề xí ạ!
b) \(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)
8 tháng 12 2015
Câu A=4
Cách giải:
\(\left(5\sqrt{3}+2\sqrt{12}-\sqrt{75}\right):\sqrt{3}\)
\(=\left(5\sqrt{3}+2\sqrt{4\cdot3}-\sqrt{25\cdot3}\right)\)\(:\sqrt{3}\)
\(=\left(5\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\)\(:\sqrt{3}\)
NW
28 tháng 6 2016
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
7 tháng 9 2020
+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)
\(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)
\(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)
\(=4\sqrt{3}\approx6,9282\)
+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)
\(=\sqrt{x-9+6\sqrt{x-9}+9}\)
\(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)
\(=\left|\sqrt{x-9}-3\right|\)
7 tháng 9 2020
\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)
13 tháng 9 2021
a, \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{17-2.3.2\sqrt{2}}-\sqrt{17+2.3.2\sqrt{2}}\)
\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}=\left|3-2\sqrt{2}\right|-\left|3+2\sqrt{2}\right|\)
\(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)
b, \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{31-2.2.3\sqrt{3}}-\sqrt{31+2.2.3\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-2\right)^2}-\sqrt{\left(3\sqrt{3}+2\right)^2}=\left|3\sqrt{3}-2\right|-\left|3\sqrt{3}+2\right|\)
\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)
7 tháng 9 2020
\(A=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)
\(C=\sqrt{6+2\sqrt{3}-2}\)
\(C=\sqrt{4+2\sqrt{3}}\)
\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
7 tháng 9 2020
1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1\)
\(=2\sqrt{2}\approx2,82843\)
2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)
\(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)
3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)
\(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)
18 tháng 7 2017
\(12-2\sqrt{35}\)
\(=\left(\sqrt{5}\right)^2+\left(\sqrt{7}\right)^2-2\sqrt{35}\)
\(=\left(\sqrt{5}+\sqrt{7}\right)^2\)
\(7+\sqrt{40}\)
\(=\left(\sqrt{5}\right)^2+\left(\sqrt{2}\right)^2+2\sqrt{10}\)
\(=\left(\sqrt{5}+\sqrt{2}\right)^2\)
3 tháng 2 2022
\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
Q
24 tháng 6 2017
a) \(\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}=\sqrt{7}-\sqrt{5}\)
b) \(\sqrt{4+\sqrt{15}}=...\)
c) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}=\left(3\sqrt{2}\right)\sqrt{\left(3+\sqrt{2}\right)^2}\\ =\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)=9-2=7\)
d) \(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}=\left(\sqrt{7}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}\\ =\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=7-5=2\)
e) \(\sqrt{7-2\sqrt{10}-\sqrt{7+2\sqrt{10}}}=\sqrt{7-2\sqrt{10}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\\ =\sqrt{7-2\sqrt{10}-\left(\sqrt{2}+\sqrt{5}\right)}=\sqrt{7-2\sqrt{10}-\sqrt{2}-\sqrt{5}}\\ =\sqrt{7-2\sqrt{10}-\sqrt{2}-\sqrt{5}}\)
f) \(\sqrt{13-\sqrt{160}+\sqrt{53}+4\sqrt{90}}=\sqrt{13-4\sqrt{10}+\sqrt{53}+12\sqrt{10}}\\ =\sqrt{13+8\sqrt{10}+\sqrt{53}}\)
\(\sqrt{12+2\sqrt{35}}=\sqrt{7+2\sqrt{7.5}+5}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\sqrt{7}+\sqrt{5}\)
\(\sqrt{12+2\sqrt{35}}=\sqrt{5}+\sqrt{7}\)