A=(4\sqrt{2}+3).\sqrt{41-24\sqrt{2}}
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ND
29 tháng 7 2018
\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}\)
\(P=\dfrac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}\)
\(P=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\)
*P/S: đã nhỡ làm câu a, câu b bạn Phùng Khánh Linh làm rồi :)
29 tháng 7 2018
\(P=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\dfrac{8\sqrt{41}}{\sqrt{41+2.2\sqrt{41}+4}+\sqrt{41-2.2\sqrt{41}+4}}=\dfrac{8\sqrt{41}}{2\sqrt{41}}=4\) \(Q=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(3-\sqrt{3}\right)+\left(2\sqrt{2}-\sqrt{6}\right)\left(3+\sqrt{3}\right)}{9-3}=\dfrac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}=\dfrac{12\sqrt{6}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
26 tháng 5 2022
\(A=\dfrac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}\cdot\dfrac{1}{\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{4}{\sqrt{3}-\sqrt{2}}=4\sqrt{3}+4\sqrt{2}\)
EL
1
12 tháng 5 2018
\(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{45-\sqrt{41}}}}:\left(\sqrt{3}-\sqrt{2}\right)\) ( đề)
\(=\frac{8\sqrt{41}}{\sqrt{41}+2-\sqrt{41}-2}:\left(\sqrt{3}-\sqrt{2}\right)\)
\(=2\sqrt{41}:\left(\sqrt{3}-\sqrt{2}\right)\)
\(=2\sqrt{123}+2\sqrt{82}\)
vậy.....................
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2020
\(A=\frac{8\sqrt{41}}{\sqrt{\sqrt{41}^2+2.2.\sqrt{41}+2^2}+\sqrt{\sqrt{41}^2-2.2.\sqrt{41}+2^2}}.\frac{1}{\sqrt{3}-\sqrt{2}}\)
\(=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}.\frac{\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\frac{8\sqrt{41}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}\left(\sqrt{3}+\sqrt{2}\right)}{2\sqrt{41}}=4\left(\sqrt{3}+\sqrt{2}\right)\)
20 tháng 7 2017
a) \(\sqrt{\dfrac{2-\sqrt{3}}{2}}+\dfrac{1-\sqrt{3}}{2}\)
= \(\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)
= \(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}+\dfrac{1-\sqrt{3}}{2}\)
= \(\dfrac{\sqrt{3}-1+1-\sqrt{3}}{2}\)
= 0
b) \(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{15}}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18+20+3+2\sqrt{54}-2\sqrt{360}-2\sqrt{60}}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{\left(\sqrt{18}-\sqrt{20}+\sqrt{3}\right)^2}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18}-2\sqrt{5}+\sqrt{3}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18}\)
Dùng lệnh tex ( đề bài )
\(A=(4\sqrt{2}+3).\sqrt{41-24\sqrt{2}}\)
\(A=\left(4\sqrt{2}+3\right).\sqrt{9+2.3.4\sqrt{2}+32}\)
\(A=\left(4\sqrt{2}+3\right).\sqrt{\left(3+4\sqrt{2}\right)^2}\)
\(A=\left(4\sqrt{2}+3\right).\left(4\sqrt{2}+3\right)\)
\(A=\left(4\sqrt{2}+3\right)^2\)