AB
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
KT
4 tháng 7 2018
a) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left(\sqrt{5}-2\right)-\left(\sqrt{5}+2\right)=-4\)
b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\frac{1}{\sqrt{2}}.\left(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1\right)=-\sqrt{2}\)
c) \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
\(=7-3\sqrt{5}-\left(7+3\sqrt{5}\right)=-6\sqrt{5}\)
1 tháng 8 2018
\(a.\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\) \(b.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\) \(c.\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
Dễ thấy \(A>0\)
\(A^2=94-42\sqrt{5}+94+42\sqrt{5}+2\sqrt{\left(94^2-42^2.5\right)}\)
\(A^2=188+2\sqrt{16}=196\)
\(\Rightarrow A=14\)
TN
17 tháng 6 2019
\(a,\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{8+2.2\sqrt{2}.5+25}-\sqrt{2-2.\sqrt{2}.3+9}\)
\(=\sqrt{\left[2\sqrt{2}+5\right]^2}-\sqrt{\left[\sqrt{2}-3\right]^2}\)
\(=2\sqrt{2}+5-\left(3-\sqrt{2}\right)\)
\(=2+\sqrt{2}\)
chúc bn học tốt
17 tháng 6 2019
a) \(\sqrt{\left(2\sqrt{2}+5\right)^2}\) \(-\) \(\sqrt{\left(3-\sqrt{2}\right)^2}\)= \(|2\sqrt{2}+5|\)\(-\)\(|3-\sqrt{2}|\)
\(=\)\(2\sqrt{2}+5-3+\sqrt{2}=2+3\sqrt{2}\)
b)\(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
HT
17 tháng 6 2019
a/ \(=\sqrt{\left(5+2\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=5+2\sqrt{2}-3+\sqrt{2}=2+3\sqrt{2}\)
b/ \(=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
NT
17 tháng 6 2019
a)
\(\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{33+2\sqrt{200}}-\sqrt{11-2\sqrt{18}}\\ =\sqrt{\sqrt{8^2}+2\sqrt{8}\sqrt{25}+5^2}-\sqrt{\sqrt{2^2}-2\sqrt{2}\sqrt{9}+3^2}\\ =\sqrt{\left(\sqrt{8}+5\right)^2}-\sqrt{\left(\sqrt{2}-3\right)^2}\\ =\left|\sqrt{8}+5\right|-\left|\sqrt{2}-3\right|\\ =\sqrt{8}+5-3+\sqrt{2}\\ =3\sqrt{2}+2\)
b)
\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\\ =\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\\ =\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\\ =7-\sqrt{45}-7-\sqrt{45}\\ =-2\sqrt{45}\)
23 tháng 9 2017
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
CB
1
17 tháng 6 2021
a) A = \(13-2\sqrt{42}=\left(\sqrt{7}-\sqrt{6}\right)^2\)
<=> \(\sqrt{A}=\sqrt{7}-\sqrt{6}\)
b) \(A=46+6\sqrt{5}=\left(\sqrt{45}+1\right)^2\)
<=> \(\sqrt{A}=\sqrt{45}+1\)
c) \(A=12-3\sqrt{15}=\dfrac{1}{2}\left(24-6\sqrt{15}\right)=\dfrac{1}{2}\left(\sqrt{15}-3\right)^2\)
<=> \(\sqrt{A}=\dfrac{1}{\sqrt{2}}\left(\sqrt{15}-3\right)\)
8 tháng 7 2019
a)\(\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{3-2\sqrt{6}+2}\)
\(=\sqrt{3-2\sqrt{2}\sqrt{3}+2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(\left|\sqrt{3}-\sqrt{2}\right|\)
B
8 tháng 7 2019
\(a,\sqrt{5-2\sqrt{6}}=\left(\sqrt{2}-\sqrt{3}\right)^2=|\sqrt{2}-\sqrt{3}|=\sqrt{3}-\sqrt{2}\)
\(b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-\left(20-10\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
\(c,\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-7\right)^2}-\sqrt{\left(3\sqrt{5}+7\right)^2}\)
\(=|3\sqrt{5}-7|-|3\sqrt{5}+7|\)
\(=7-3\sqrt{5}-3\sqrt{5}-7\)
\(=-6\sqrt{5}\)
AB
0
\(\sqrt{a+42}-\sqrt{a-42}=44\)
<=>\(\frac{84}{\sqrt{a+42}-\sqrt{a-42}}=44\)
<=>\(\sqrt{a+42}-\sqrt{a-42}=\frac{21}{11}\)