$\frac{2 \sqrt{3+ \sqrt{5}-\sqrt{13+ \sqrt{48}}}}{\sqrt{6}+ \sqrt{2}}$
Những câu hỏi liên quan
Rút gọnAfrac{frac{sqrt{2+sqrt{3}}}{2}}{frac{sqrt{2+sqrt{3}}}{2}-frac{2}{sqrt{6}}+frac{sqrt{2+sqrt{3}}}{2sqrt{3}}}B frac{2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}}{sqrt{6}+sqrt{2}}C frac{left(5+2sqrt{6}right)left(49-20sqrt{16}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}Dsqrt{4+sqrt{5+sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}Mn lm đc câu nào thì đc, hết càng tốt 3
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Rút gọn
A=\(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
B= \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
C= \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{16}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{5+\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Mn lm đc câu nào thì đc, hết càng tốt <3
1. Tính:
a) \(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
b) \(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
c) \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
d) \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
Tính : \(\frac{\sqrt{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}}{\sqrt{6}+\sqrt{2}}\)
Thực hiện các phép tính sau:
a) \(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
b) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
c) \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
Mai em kiểm tra! Giúp em vs! em cảm ơn nhiều!!
Rút gọn
1) \(B=\sqrt{\sqrt{7}+6+\sqrt{13-2\sqrt{64-6\sqrt{7}}}}\)
2) \(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
3) \(D=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Tính \(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Thực hiện phép tính sau
a, \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
b, \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
c, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
d, \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)
\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)
c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)
\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)
\(=\dfrac{\sqrt{1\cdot4}}{2}\)
\(=\dfrac{2}{2}\)
\(=1\)
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12 tháng 7 2021 lúc 10:06
có ai biết giải ko giải hộ mình mấy bài này với ( giải chi tiết hộ mình nhé)1, 2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}2, sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}3, sqrt{4+sqrt{5sqrt{3+}5sqrt{48-10sqrt{7+4sqrt{3}}}}}4, sqrt{30-2sqrt{16+6sqrt{11+4sqrt{4-2sqrt{3}}}}}5, dfrac{left(5sqrt{3}+sqrt{50}right)left(5-sqrt{24}right)}{sqrt{75}-5sqrt{2}}6, sqrt{4+sqrt{8}.sqrt{2+sqrt{2}}}.sqrt{2-sqrt{2+sqrt{2}}}7, sqrt{8sqrt{3}-2sqrt{25sqrt{12}+4sqrt{192}}}
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có ai biết giải ko giải hộ mình mấy bài này với ( giải chi tiết hộ mình nhé)
1, \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
2, \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
3, \(\sqrt{4+\sqrt{5\sqrt{3+}5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
4, \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)
5, \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
6, \(\sqrt{4+\sqrt{8}.\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
7, \(\sqrt{8\sqrt{3}-2\sqrt{25\sqrt{12}+4\sqrt{192}}}\)
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
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4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)
\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)
\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)
\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)
\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)
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5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)
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Xem thêm câu trả lời
chứng minh đẳng thức
\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}=1\)
Xét tử \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}=2\sqrt{3+\sqrt{3}-1}\)
\(=2\sqrt{2+\sqrt{3}}=\frac{2\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{2\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{2\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
Suy ra VT = VP = 1
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Đáp án là : VT = VP = 1
vhuv bn hoc gioi tk mk nha
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ai hâm mộ dĩnh tỷ nhớ kết bạn với mình nhé
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