\(\sqrt{4+5\sqrt{3+5}\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
Giúp mk giải bài toán này nha !
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
DV
0
TN
9 tháng 9 2017
CÁi này easy mà .-.
\(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
\(\Leftrightarrow\frac{\frac{\left(7-x\right)-\left(x-5\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)
\(\Leftrightarrow\frac{\frac{-2\left(x-6\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{\frac{-2}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+1\right)=0\)
\(\Rightarrow x-6=0\Rightarrow x=6\)
29 tháng 1 2022
a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
BG
2
14 tháng 7 2021
Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{25}=5\)
14 tháng 7 2021
\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{4+2.2\sqrt{3+\left(\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-20+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{28+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}\right)^2}+2.5.\sqrt{3}+5^2}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)^2}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{3+10}}}\)
=\(10\sqrt{3+10}\)
=\(\sqrt{10\left(\sqrt{3+1}\right)}\)
24 tháng 8 2021
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
\(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+3}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{48-10\left(2+\sqrt{3}\right)}=\sqrt{48-20-10\sqrt{3}}=\sqrt{28-10\sqrt{3}}=\sqrt{5^2-2.5.\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
mk thay de no cu the nao ay