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1 tháng 8 2018
Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)
\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)
\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)
Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)
\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)
\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)
Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)
\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)
\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)
Chúc bạn học tốt
1 tháng 8 2018
4 , Ta có :
\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)
\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)
\(=\sqrt{\left(\sqrt{3}-6\right)^2}-\sqrt{\left(\sqrt{3}+6\right)^2}\)
\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)
\(=6-\sqrt{3}-\sqrt{3}-6\)
\(=-2\sqrt{3}\)
25 tháng 7 2018
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
25 tháng 7 2018
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
22 tháng 9 2018
\(6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)
\(\Rightarrow\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{6+\sqrt{6+\sqrt{6+3}}}\)
\(=\sqrt{6+\sqrt{6+3}}\)\(=\sqrt{6+3}\)\(=3\)
\(12>9\Rightarrow\sqrt{12}>\sqrt{9}=3\)
\(\Rightarrow\sqrt{12+\sqrt{12+\sqrt{12}}}>\sqrt{12}>3\)
\(\Rightarrow\sqrt{12+\sqrt{12+\sqrt{12}}}>\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\)
DM
0
12 tháng 7 2018
\(\text{a) }\sqrt{16-8\sqrt{3}}=\sqrt{12+4-4\sqrt{12}}=\sqrt{\left(\sqrt{12}-2\right)^2}=\sqrt{12}-2\)
\(\text{b) }\sqrt{38+12\sqrt{2}}=\sqrt{36+2+12\sqrt{2}}=\sqrt{\left(6+\sqrt{2}\right)^2}=6+\sqrt{2}\)
\(\text{c) }\sqrt{22+12\sqrt{2}}=\sqrt{18+2+4\sqrt{18}}=\sqrt{\left(\sqrt{18}+\sqrt{2}\right)^2}=3\sqrt{2}+\sqrt{2}=4\sqrt{2}\)
\(\text{d) }\sqrt{17-12\sqrt{2}}=\sqrt{9+8-6\sqrt{8}}=\sqrt{\left(3-\sqrt{8}\right)^2}=3-\sqrt{8}\)
\(\text{e) }\sqrt{20-10\sqrt{3}}=\sqrt{15+5-2\sqrt{75}}=\sqrt{\left(\sqrt{15}-\sqrt{5}\right)^2}=\sqrt{15}-\sqrt{5}\)
13 tháng 2 2022
Đặt \(\sqrt{12}=a;\sqrt{13}=b\)
Theo đề, ta có:
\(\dfrac{a^2}{b}+\dfrac{b^2}{a}>a+b\)
\(\Leftrightarrow a^2+b^2-a^2-2ab-b^2>0\)
\(\Leftrightarrow2ab< 0\)(đúng)
3 tháng 8 2019
\(\sqrt{5-\sqrt{21}}=\sqrt{\frac{1}{2}}.\sqrt{10-2\sqrt{21}}=\sqrt{\frac{1}{2}}.\sqrt{3-2\sqrt{3}\sqrt{7}+7}=\sqrt{\frac{1}{2}}\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{\frac{1}{2}}.\sqrt{7}-\sqrt{\frac{1}{2}}.\sqrt{3}=\sqrt{3,5}-\sqrt{1,5}\)
\(\sqrt{7+3\sqrt{5}}=\sqrt{\frac{1}{2}\left(14+2.3\sqrt{5}\right)}=\sqrt{\frac{1}{2}\left(5+2.3\sqrt{5}+3^2\right)}=\sqrt{\frac{1}{2}\left(3+\sqrt{5}\right)^2}=\sqrt{\frac{1}{2}}\left(3+\sqrt{5}\right)=\sqrt{4,5}+\sqrt{2,5}\)
\(\sqrt{49+5\sqrt{96}}=\sqrt{49+2.2.5\sqrt{6}}=\sqrt{2^2.6+2.2.5\sqrt{6}+5^2}=\sqrt{\left(5+2\sqrt{6}\right)^2}=5+2\sqrt{6}\)
3 tháng 8 2019
\(\sqrt{5-\sqrt{21}}=\frac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7\cdot3}+3}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)
\(\sqrt{7+3\sqrt{5}}=\frac{\sqrt{14+6\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{9+2\cdot3\sqrt{5}+4}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}=\frac{3+\sqrt{5}}{\sqrt{2}}\)
\(\sqrt{49+5\sqrt{96}}=\sqrt{49+5\sqrt{4\cdot24}}=\sqrt{25+2\cdot5\sqrt{24}+24}=\sqrt{\left(5+\sqrt{24}\right)^2}=5+\sqrt{24}\)
\(\sqrt{51-7\sqrt{8}}=\sqrt{51-7\sqrt{2^2\cdot2}}=\sqrt{49-2\cdot7\sqrt{2}+2}=\sqrt{\left(7+\sqrt{2}\right)^2}=7+\sqrt{2}\)
\(\sqrt{28+5\sqrt{12}}=\sqrt{28+5\sqrt{2^2\cdot3}}=\sqrt{25+2\cdot5\sqrt{3}+3}=\sqrt{\left(5+\sqrt{3}\right)^2}=5+\sqrt{3}\)
\(\sqrt{12-3\sqrt{12}}=\sqrt{12-3\sqrt{2^2\cdot3}}=\sqrt{9-2\cdot3\sqrt{3}+3}=\sqrt{\left(3+\sqrt{3}\right)^2}=3+\sqrt{3}=\sqrt{3}\left(\sqrt{3}+1\right)\)
Chúc bạn học tốt nha
.
10 tháng 7 2018
\(\sqrt{\left(4-\sqrt{15}\right)^2}=\left|4-\sqrt{15}\right|=4-\sqrt{15}\)
\(\Rightarrow\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}=4-\sqrt{15}+\sqrt{15}=4\)
\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
\(\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
\(\Rightarrow\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}=2-\sqrt{3}+\sqrt{3}-1=1\)
Đặt \(a=\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}\left(a>0\right)\)
\(\Rightarrow a^2=12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}\)
\(\Rightarrow a^2-a=\)\(12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12}}}}\)\(-\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}-12=0\)
\(\Rightarrow a^2-a-12=0\)
\(\Leftrightarrow\left(a+3\right)\left(a-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-3\left(ktm\right)\\a=4\left(tm\right)\end{matrix}\right.\)
Vậy \(\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}=4\)
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