giải giúp em với:\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4+\sqrt{3}}}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
AT
12 tháng 7 2021
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
AT
12 tháng 7 2021
1) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+\sqrt{4.12}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}\right)^2+2.\sqrt{12}+1^2}}=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}=\sqrt{5-\left|\sqrt{4.3}+1\right|}\)
\(=\sqrt{5-\left(2\sqrt{3}+1\right)}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{3}-1}=2\sqrt{2+\sqrt{3}}\)
\(=2\sqrt{\dfrac{4+2\sqrt{3}}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}\)
\(=2.\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
2) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{3}-1\) (như trên)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
NV
30 tháng 6 2016
a/ A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
Vậy A = 1
NV
17 tháng 9 2015
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\left(5-\sqrt{3}\right)}}=\sqrt{5\sqrt{3}+\sqrt{25-5\sqrt{3}}}\)
Trần Đức Thắng lm nốt đi
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
NN
2
Y
7 tháng 7 2023
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{5^2-2.5.\sqrt{3}+\sqrt{3^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}\)
\(=\sqrt{9}\\ =3\)
DA
7 tháng 7 2023
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.|2+\sqrt{3}|}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.\left(2+\sqrt{3}\right)}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5.|5-\sqrt{3}|}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5.\left(5-\sqrt{3}\right)}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
= \(\sqrt{4+\sqrt{25}}\)
= \(\sqrt{4+5}\)
= \(\sqrt{9}\)
= \(3\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{5^2-2\cdot5\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
25 tháng 7 2023
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}=\sqrt{25}=5\)
UK
10 tháng 9 2017
:VV Đề vậy thì khó thiệt :V, cho sửa đề tí
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Tính từ trong ra :V
\(7+4\sqrt{3}=7+2.2\sqrt{3}=2^2+2.2\sqrt{3}+\sqrt{3}^2=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{7+4\sqrt{3}}=2+\sqrt{3}\)
\(\sqrt{48-10\left(2+\sqrt{3}\right)}=\sqrt{28-10\sqrt{3}}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
\(\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}=\sqrt{5\sqrt{3}+25-5\sqrt{3}}=\sqrt{25}=5\)
\(\sqrt{4+5}=\sqrt{9}=3\)
Vậy cái lùm xùm đó bằng 3
NT
1
TT
23 tháng 7 2016
\(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+3}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{48-10\left(2+\sqrt{3}\right)}=\sqrt{48-20-10\sqrt{3}}=\sqrt{28-10\sqrt{3}}=\sqrt{5^2-2.5.\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
mk thay de no cu the nao ay