1) \(\sqrt{7-2\sqrt{10}}\) - \(\sqrt{7+2\sqrt{10}}\)
2) \(\sqrt{4-2\sqrt{3}}\) + \(\sqrt{4+2\sqrt{3}}\)
3) \(\sqrt{6-4\sqrt{2}}\) + \(\sqrt{22-12\sqrt{2}}\)
Những câu hỏi liên quan
a. P (3+sqrt{2}+sqrt{6})(sqrt{6-3sqrt{3}})
b. A(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}): (sqrt{6}+11)
c. B frac{sqrt{8-2sqrt{12}}}{sqrt{3}-1}-sqrt{8}
d. C sqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}-sqrt{2}
đ. Dfrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}+2sqrt{3}}}
e. E sqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
ê. G sqrt{4+5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}
g. Hfrac{2sqrt{4+sqrt{5+21+sqrt{80}}}}{sqrt{10}-sqrt{2}}
i. Isqrt{frac{4-sqrt{7}}{4+sqrt{7...
Đọc tiếp
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
1. sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}+sqrt{8}
2. dfrac{sqrt{3-2sqrt{3}}}{sqrt{17-12sqrt{2}}}-dfrac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
3.sqrt{7+2sqrt{6}}-sqrt{left(sqrt{6-1}right)^2}
4sqrt{5-2sqrt{6}}-sqrt{5+sqrt{24}}
5.sqrt{4sqrt{5+sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}}
6.sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
Đọc tiếp
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6-1}\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
1. sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}+sqrt{8}
2. dfrac{sqrt{3-2sqrt{3}}}{sqrt{17-12sqrt{2}}}-dfrac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
3.sqrt{7+2sqrt{6}}-sqrt{left(sqrt{6}-1right)^2}
4sqrt{5-2sqrt{6}}-sqrt{5+sqrt{24}}
5.sqrt{4sqrt{5+sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}}
6.sqrt{6+2sqrt{5-sqrt{13+sqrt{48}}}}
Đọc tiếp
1. \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
2. \(\dfrac{\sqrt{3-2\sqrt{3}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
3.\(\sqrt{7+2\sqrt{6}}-\sqrt{\left(\sqrt{6}-1\right)^2}\)
4\(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{24}}\)
5.\(\sqrt{4\sqrt{5+\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
6.\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Bài 1: Tính
\(\sqrt{3+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\\ \sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}\\ \sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}\\ \sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}\\ \sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\\ \sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
1.
$\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3}}-\sqrt{3+1-2\sqrt{3}}$
$=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}$
$=|\sqrt{3}+1|-|\sqrt{3}-1|=2$
2.
\(\sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}=\sqrt{12+6\sqrt{3}+\sqrt{9+3-2\sqrt{9.3}}}=\sqrt{12+6\sqrt{3}+\sqrt{(3-\sqrt{3})^2}}\)
\(=\sqrt{12+6\sqrt{3}+3-\sqrt{3}}=\sqrt{15+5\sqrt{3}}\)
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3.
\(\sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{9-4\sqrt{2}+\sqrt{8+1+2\sqrt{8.1}}}\)
\(=\sqrt{9-4\sqrt{2}+\sqrt{2\sqrt{2}+1)^2}}=\sqrt{9-4\sqrt{2}+2\sqrt{2}+1}=\sqrt{10-2\sqrt{2}}\)
4.
\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8+1-2\sqrt{8.1}}}}\)
\(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{(\sqrt{8}-1)^2}}}\) \(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8}-1}}=\sqrt{\sqrt{2}+2+\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{\sqrt{2}+2+\sqrt{(2+1+2\sqrt{2}}}=\sqrt{\sqrt{2}+2+\sqrt{(\sqrt{2}+1)^2}}=\sqrt{\sqrt{2}+2+\sqrt{2}+1}\)
\(=\sqrt{3+2\sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1\)
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5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
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Xem thêm câu trả lời
Tính giá trị các biểu thức:
a.\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\sqrt{3}\)
b.\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
c.\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
d.\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
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a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
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d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)
\(=19\sqrt{3\sqrt{2}}\)
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\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)\(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
Đọc tiếp
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Tính:
1) \(\sqrt{4-2\sqrt{3}}\)
2) \(\sqrt{5+2\sqrt{6}}\)
3) \(\sqrt{7-2\sqrt{10}}\)
4) \(\sqrt{14-6\sqrt{6}}\)
5) \(\sqrt{8+2\sqrt{15}}\)
6) \(\sqrt{10-2\sqrt{21}}\)
7) \(\sqrt{11+2\sqrt{18}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
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Rút gọn biểu thức:
a) \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
b) \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)
c) \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
d) \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)
\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)
\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)
\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}\)
\(=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)
\(=\sqrt{\sqrt{3}+5-\sqrt{3}}=\sqrt{5}\)
\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)
\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)
\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)
\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)
\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)
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a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)
\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{\sqrt{3}+5-\sqrt{3}}\)
\(=\sqrt{5}\)
d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)
\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)
\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)
\(=\sqrt{11+6\sqrt{2}}\)
\(=3+\sqrt{2}\)
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