\(H=\sqrt{2+\sqrt{3}}\cdot\sqrt{14-5\sqrt{3}}+\sqrt{2}\)
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19 tháng 10 2021
\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}\cdot\sqrt{2}\left(3+\sqrt{7}\right)\\ =\sqrt{\dfrac{2\left(3+\sqrt{7}\right)^2}{8+3\sqrt{7}}}=\sqrt{\dfrac{32+12\sqrt{7}}{8+3\sqrt{7}}}\\ =\sqrt{\dfrac{4\left(8+3\sqrt{7}\right)}{8+3\sqrt{7}}}=\sqrt{4}=2\)
27 tháng 10 2019
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
L
22 tháng 7 2021
a)\(\sqrt{3\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{2}+2\sqrt{3}}\)
= \(\sqrt{18-12}\)
= \(\sqrt{6}\)
b) \(\sqrt{2+2\sqrt{2-\sqrt{2}}}.\sqrt{2-2\sqrt{2-\sqrt{2}}}\)
= \(\sqrt{4-4\left(\sqrt{2-\sqrt{2}}\right)^2}\)
= \(\sqrt{4-4.\left(2-4\sqrt{2}+2\right)}\)
= \(\sqrt{4-8+16\sqrt{2}-8}\)
= \(\sqrt{-12+16\sqrt{2}}\)
c)
\(\left(\sqrt{2}-\sqrt{7}\right).\sqrt{9+2\sqrt{14}}\)
= \(\left(\sqrt{2}-\sqrt{7}\right).\left(2+2\sqrt{7}.\sqrt{2}+7\right)\)
= \(\left(\sqrt{2}-\sqrt{7}\right).\left(\sqrt{2}+\sqrt{7}\right)^2\)
= \(\left(4-7\right).\left(\sqrt{2}+\sqrt{7}\right)\)
= \(-3.\left(\sqrt{2}+\sqrt{7}\right)\)
15 tháng 7 2016
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}}=\sqrt{\left(3^2\right)-\left(\sqrt{5+2\sqrt{3}}\right)^2}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)
\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
15 tháng 7 2016
\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
MD
13 tháng 6 2017
Câu 2b đề là tìm x chứ nhỉ???
b) \(\sqrt{x^2-4}+\sqrt{x-2}=0\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x-2}\ge0\end{matrix}\right.\)
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\pm2\\x=2\end{matrix}\right.\) <=> x = 2
Vậy x = 2
13 tháng 6 2017
bài 2 câu b) đề sai rồi bạn
còn bài 1 câu b) mình cảm thấy sai sai
25 tháng 7 2022
a: \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(=\dfrac{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}}{\sqrt{2}}\)
\(=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
2 tháng 2 2022
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
9 tháng 7 2022
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
AH
Akai Haruma
Giáo viên
23 tháng 11 2018
Câu 1:
Có: \(8-4\sqrt{3}=8-2\sqrt{12}=6+2-2\sqrt{6.2}=(\sqrt{6}-\sqrt{2})^2\)
\(\Rightarrow \sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}\)
Do đó:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}\)
\(=\sqrt{(\sqrt{6})^2-(\sqrt{2})^2}=\sqrt{6-2}=2\)
AH
Akai Haruma
Giáo viên
23 tháng 11 2018
Câu 2:
\(16-5\sqrt{7}=\frac{32-10\sqrt{7}}{2}=\frac{32-2\sqrt{175}}{2}=\frac{25+7-2\sqrt{25.7}}{2}=\frac{(5-\sqrt{7})^2}{2}\)
\(\Rightarrow \sqrt{16-5\sqrt{7}}=\frac{5-\sqrt{7}}{\sqrt{2}}\)
Do đó:
\(\sqrt{16-5\sqrt{7}}(5\sqrt{2}+\sqrt{14})+\frac{6}{3+\sqrt{10}}=\frac{5-\sqrt{7}}{\sqrt{2}}.\sqrt{2}(5+\sqrt{7})+\frac{6(3-\sqrt{10})}{(3+\sqrt{10})(3-\sqrt{10})}\)
\(=(5-\sqrt{7})(5+\sqrt{7})+\frac{18-6\sqrt{10}}{3^2-10}=25-7+(-18+6\sqrt{10})\)
\(=6\sqrt{10}\)
Ai giả đc giúp mk với
\(H=\sqrt{2+\sqrt{3}}.\sqrt{14-5\sqrt{3}}+\sqrt{2}.\)
\(2.H=\sqrt{2\left(2+\sqrt{3}\right)}.\sqrt{2\left(14-5\sqrt{3}\right)}+2\sqrt{2}\)
\(2H=\sqrt{3+2\sqrt{3}+1}.\sqrt{25-2.5.\sqrt{3}+3}+2\sqrt{2}\)
\(2H=\sqrt{\left(\sqrt{3}+1\right)^2}.\sqrt{\left(5-\sqrt{3}\right)^2}+2\sqrt{2}\)
\(2H=\left(\sqrt{3}+1\right)\left(5-\sqrt{3}\right)+2\sqrt{2}\)
\(2H=5\sqrt{3}-3+5-\sqrt{3}+2\sqrt{2}\)
\(2H=2+4\sqrt{3}+2\sqrt{2}\)
\(H=1+2\sqrt{3}+\sqrt{2}.\)
(P/S : đừng k cho câu trả lời này nhé)