Tính
G.\(\sqrt{7-2\sqrt{6}}\)
H.\(\sqrt{13-4\sqrt{3}}\)
I. \(\sqrt{7-4\sqrt{3}}\)\(-2\)
J.\(\sqrt{15-6\sqrt{6}}\)+\(\sqrt{33-12\sqrt{6}}\)
Những câu hỏi liên quan
Tính:
1) \(\sqrt{14-2\sqrt{33}}\)
2) \(\sqrt{12-2\sqrt{35}}\)
3) \(\sqrt{16-2\sqrt{55}}\)
4) \(\sqrt{14-6\sqrt{5}}\)
5) \(\sqrt{17-12\sqrt{2}}\)
6) \(\sqrt{27-12\sqrt{5}}\)
7) \(\sqrt{4+\sqrt{15}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
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Tính:
a) left(2sqrt{5}-sqrt{7}right).left(2sqrt{5}+sqrt{7}right)
b)left(sqrt{5-2sqrt{6}}+sqrt{2}right).sqrt{3}
c)sqrt{7-4sqrt{3}}+sqrt{7+4sqrt{3}}
d)sqrt{15-6sqrt{6}}+sqrt{33-12sqrt{6}}
e)sqrt{3+sqrt{5}}+sqrt{3-sqrt{5}}
g)sqrt{8-2sqrt{15}}-sqrt{23-4sqrt{15}}
Đọc tiếp
Tính:
a) \(\left(2\sqrt{5}-\sqrt{7}\right).\left(2\sqrt{5}+\sqrt{7}\right)\)
b)\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right).\sqrt{3}\)
c)\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
d)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e)\(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
g)\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
a)
\((2\sqrt{5}-\sqrt{7})(2\sqrt{5}+\sqrt{7})=(2\sqrt{5})^2-(\sqrt{7})^2=13\)
b)
\((\sqrt{5-2\sqrt{6}}+\sqrt{2})\sqrt{3}=(\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2})\sqrt{3}\)
\(=(\sqrt{(\sqrt{3}-\sqrt{2})^2}+\sqrt{2})\sqrt{3}=(\sqrt{3}-\sqrt{2}+\sqrt{2})\sqrt{3}=\sqrt{3}.\sqrt{3}=3\)
c)
\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)
\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}=2-\sqrt{3}+2+\sqrt{3}=4\)
d)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{3^2+6-2.3\sqrt{6}}+\sqrt{9+24-2\sqrt{9.24}}\)
\(=\sqrt{(3-\sqrt{6})^2}+\sqrt{(\sqrt{24}-3)^2}=3-\sqrt{6}+\sqrt{24}-3\)
\(=\sqrt{6}\)
e)
\(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=\sqrt{\frac{6+2\sqrt{5}}{2}}+\sqrt{\frac{6-2\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5+1+2\sqrt{5.1}}{2}}+\sqrt{\frac{5+1-2\sqrt{5.1}}{2}}=\sqrt{\frac{(\sqrt{5}+1)^2}{2}}+\sqrt{\frac{(\sqrt{5}-1)^2}{2}}\)
\(=\frac{\sqrt{5}+1}{\sqrt{2}}+\frac{\sqrt{5}-1}{\sqrt{2}}=\sqrt{10}\)
g)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{20+3-2\sqrt{20.3}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{20}-\sqrt{3})^2}\)
\(=\sqrt{5}-\sqrt{3}-(\sqrt{20}-\sqrt{3})=\sqrt{5}-\sqrt{20}=-\sqrt{5}\)
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=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Bài 1: TínhAsqrt{46-6sqrt{5}}-sqrt{29-12sqrt{5}}Bsqrt{13-sqrt{160}-sqrt{53+4sqrt{90}}}Csqrt{15-6sqrt{6}}+sqrt{35-12sqrt{6}}Dsqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}E sqrt{4-sqrt{7}}+sqrt{4+sqrt{7}}F sqrt{3+sqrt{11+6sqrt{2}}}-sqrt{5+2sqrt{6}}Gsqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}Bài 2: so sánha) sqrt{24}+sqrt{45} và 12b) sqrt{37}-sqrt{15} và 2c) sqrt{16} và sqrt{15}timessqrt{17}d) 8 và sqrt{15}+sqrt{17}
Đọc tiếp
Bài 1: Tính
A=\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
B=\(\sqrt{13-\sqrt{160}-\sqrt{53+4\sqrt{90}}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
D=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
E= \(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
F= \(\sqrt{3+\sqrt{11+6\sqrt{2}}}-\sqrt{5+2\sqrt{6}}\)
G=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
Bài 2: so sánh
a) \(\sqrt{24}+\sqrt{45}\) và 12
b) \(\sqrt{37}-\sqrt{15}\) và 2
c) \(\sqrt{16}\) và \(\sqrt{15}\times\sqrt{17}\)
d) 8 và \(\sqrt{15}+\sqrt{17}\)
Bài 2 :
a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)
b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)
c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)
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Thực hiện phép tínha) left(2-sqrt{3}right)^2-sqrt{4-2sqrt{3}}+sqrt{12}b)frac{2sqrt{12}-sqrt{6}}{2sqrt{6}-sqrt{3}}+frac{10+sqrt{5}}{2sqrt{15}+sqrt{3}}c)left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right):left(sqrt{2}+sqrt{3}right)d)sqrt{10-4sqrt{6}}+sqrt{33-12sqrt{6}}e)sqrt{16-6sqrt{7}}+sqrt{32-8sqrt{7}}f)sqrt{28-16sqrt{3}}+sqrt{73-40sqrt{3}}
Đọc tiếp
Thực hiện phép tính
a) \(\left(2-\sqrt{3}\right)^2-\sqrt{4-2\sqrt{3}}+\sqrt{12}\)
b)\(\frac{2\sqrt{12}-\sqrt{6}}{2\sqrt{6}-\sqrt{3}}+\frac{10+\sqrt{5}}{2\sqrt{15}+\sqrt{3}}\)
c)\(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right):\left(\sqrt{2}+\sqrt{3}\right)\)
d)\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e)\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
f)\(\sqrt{28-16\sqrt{3}}+\sqrt{73-40\sqrt{3}}\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
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\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
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\(\sqrt{28-16\sqrt{3}}+\sqrt{73-40\sqrt{3}}\)
\(=\sqrt{4^2-2.4.2\sqrt{3}+\left(2\sqrt{3}\right)^2}+\sqrt{5^2-2.5.4\sqrt{3}+\left(4\sqrt{3}\right)^2}\)
\(=\sqrt{\left(4-2\sqrt{3}\right)^2}+\sqrt{\left(5-4\sqrt{3}\right)^2}\)
\(=4-2\sqrt{3}-\left(5-4\sqrt{3}\right)\)
\(=4-2\sqrt{3}-5+4\sqrt{3}\)
\(=-1+2\sqrt{3}\)
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Xem thêm câu trả lời
Rút gọn
Hleft(sqrt{3}-1right)sqrt{6+2sqrt{2}.sqrt{3-sqrt{sqrt{2}+sqrt{12}+sqrt{18-sqrt{128}}}}}
Ffrac{left(5+2sqrt{6}right)left(49-20sqrt{6}right)sqrt{5-2sqrt{6}}}{9sqrt{3}-11sqrt{2}}
Gsqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
Efrac{2sqrt{3+sqrt{5-13+sqrt{48}}}}{sqrt{6}+sqrt{2}}
Dsqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}
Zsqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10-2sqrt{5}}}
Đọc tiếp
Rút gọn
H=\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
F=\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
G=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\frac{2\sqrt{3+\sqrt{5-13+\sqrt{48}}}}{\sqrt{6}+\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
Z=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
Đề thiếu nha:
\(E=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)(vì \(\sqrt{3}>1\))
\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
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\(D=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(\Rightarrow D\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
\(\Rightarrow D=\frac{2}{\sqrt{2}}=\sqrt{2}\)
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\(H=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-8\sqrt{2}+2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)(vì \(4>\sqrt{2}\))
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3+2\sqrt{3}+1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{6}+2\sqrt{2}}\)
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Xem thêm câu trả lời
a : \(\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
b : \(\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
c : \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
b) \(\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
c) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
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\(a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}\)
\(=\left|\sqrt{2}-1\right|+2-\sqrt{2}\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
\(---\)
\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}\)
\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
\(---\)
\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
#\(Toru\)
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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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Asqrt{19-3sqrt{ }40}-sqrt{19+3sqrt{ }40}
Bsqrt{21-6sqrt{ }6} +sqrt{9+2sqrt{ }18} -2sqrt{6+3sqrt{ }3}
Csqrt{6+2sqrt{ }2sqrt{ }3-sqrt{ }4+2sqrt{ }3}
Dsqrt{4+sqrt{ }15}-sqrt{7-3sqrt{ }5}
Esqrt{2+sqrt{ }3}+sqrt{2-sqrt{ }3}
Fsqrt{12-3sqrt{ }7}-sqrt{12+3sqrt{ }7}
G(3sqrt{2}+sqrt{6}).sqrt{6-3sqrt{ }3}
Hsqrt{9-4sqrt{ }5}-sqrt{14-6sqrt{ }5}
Isqrt{9-4sqrt{ }2}-sqrt{13-4sqrt{ }3}
Đọc tiếp
A=\(\sqrt{19-3\sqrt{ }40}\)-\(\sqrt{19+3\sqrt{ }40}\)
B=\(\sqrt{21-6\sqrt{ }6}\) +\(\sqrt{9+2\sqrt{ }18}\) -2\(\sqrt{6+3\sqrt{ }3}\)
C=\(\sqrt{6+2\sqrt{ }2\sqrt{ }3-\sqrt{ }4+2\sqrt{ }3}\)
D=\(\sqrt{4+\sqrt{ }15}\)-\(\sqrt{7-3\sqrt{ }5}\)
E=\(\sqrt{2+\sqrt{ }3}\)+\(\sqrt{2-\sqrt{ }3}\)
F=\(\sqrt{12-3\sqrt{ }7}\)-\(\sqrt{12+3\sqrt{ }7}\)
G=(3\(\sqrt{2}\)+\(\sqrt{6}\)).\(\sqrt{6-3\sqrt{ }3}\)
H=\(\sqrt{9-4\sqrt{ }5}-\sqrt{14-6\sqrt{ }5}\)
I=\(\sqrt{9-4\sqrt{ }2}\)-\(\sqrt{13-4\sqrt{ }3}\)
\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)
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\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)
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https://i.imgur.com/bPXZh7R.jpg