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-...

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Tứ Diệp Thảo

17 tháng 5 2020 lúc 10:51

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=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

Các câu hỏi tương tự

Dương Thanh Ngân

8 tháng 9 2020 lúc 23:02

Trục căn ở mẫu:

\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)

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Lớp 9 Toán Violympic toán 9

Tứ Diệp Thảo

17 tháng 5 2020 lúc 10:24

a. Pfrac{sqrt{4-2sqrt{3}}}{sqrt{6}-sqrt{2}}+sqrt{6+2sqrt{5}-sqrt{29-12sqrt{5}}} b.P (frac{2}{sqrt{3}-1}-frac{52}{3sqrt{3}-1}+frac{12}{3-sqrt{3}}) ( 5+sqrt{27}) c. P (frac{2+sqrt{2}}{sqrt{2}+1}+1)(frac{2-sqrt{2}}{sqrt{2}-1}-1) d. Psqrt{9+sqrt{17}}-sqrt{9-sqrt{17}}-sqrt{2} đ. P(2+sqrt{4+sqrt{6+2sqrt{5}}} )(sqrt{10}-sqrt{2} ) e. P frac{sqrt{2}+sqrt{3}+sqrt{6}+sqrt{8}+4}{sqrt{2}+sqrt{3}+sqrt{4}} ê. P sqrt{8+sqrt{8}+sqrt{20}+sqrt{40}} g. G frac{2sqrt{3-sqrt{3+sqrt{13+sqrt{48}}}}}{sqrt{6}-sqrt{...

Đọc tiếp

a. P=\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}+\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)

b.P= (\(\frac{2}{\sqrt{3}-1}-\frac{52}{3\sqrt{3}-1}+\frac{12}{3-\sqrt{3}}\)) ( 5+\(\sqrt{27}\))

c. P= (\(\frac{2+\sqrt{2}}{\sqrt{2}+1}+1\))(\(\frac{2-\sqrt{2}}{\sqrt{2}-1}-1\))

d. P=\(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}\)

đ. P=(2+\(\sqrt{4+\sqrt{6+2\sqrt{5}}}\) )(\(\sqrt{10}-\sqrt{2}\) )

e. P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

ê. P= \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

g. G= \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

h. H=\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)

i. I= \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

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Lớp 9 Toán Violympic toán 9

Tdq_S.Coups

19 tháng 9 2019 lúc 10:13

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=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)

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Lớp 9 Toán Violympic toán 9

Nguyễn Thị Kim chung

1 tháng 7 2019 lúc 6:36

1,Rút gọn:

a, \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+2}\)

b,\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)

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Lớp 9 Toán Violympic toán 9

Ngô Hoàng Anh

10 tháng 7 2019 lúc 18:16

Rút gọn a, frac{2sqrt{3-1}}{sqrt{15}}-frac{2-sqrt{5}}{sqrt{3}}-frac{4sqrt{15}-10sqrt{3}}{15} b, left(sqrt{a}-frac{1}{sqrt{a}}right)left(frac{sqrt{a}+1}{sqrt{a-1}}+frac{sqrt{a-1}}{sqrt{a}+1}right) c, sqrt{4+sqrt{7}-sqrt{4-sqrt{7}}} d, 6+2sqrt{2}.3-sqrt{4+sqrt{2sqrt{3}}} e, sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}} Help me !!!

Đọc tiếp

Rút gọn

a, \(\frac{2\sqrt{3-1}}{\sqrt{15}}-\frac{2-\sqrt{5}}{\sqrt{3}}-\frac{4\sqrt{15}-10\sqrt{3}}{15}\)

b, \(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a-1}}+\frac{\sqrt{a-1}}{\sqrt{a}+1}\right)\)

c, \(\sqrt{4+\sqrt{7}-\sqrt{4-\sqrt{7}}}\)

d, \(6+2\sqrt{2}.3-\sqrt{4+\sqrt{2\sqrt{3}}}\)

e, \(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)

Help me !!!

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Lớp 9 Toán Violympic toán 9

Tdq_S.Coups

14 tháng 9 2019 lúc 13:48

❤ Tính: a) sqrt{5-sqrt{21}}-sqrt{5+sqrt{21}} b)sqrt{13-sqrt{160}}-sqrt{53+4sqrt{90}} c)sqrt{7+sqrt{24}}+sqrt{31-sqrt{600}} d)sqrt{28-sqrt{300}}+sqrt{4-sqrt{12}} e)sqrt{7-sqrt{40}}-sqrt{5-sqrt{24}}-sqrt{6-sqrt{20}} f)sqrt{48-10sqrt{7+sqrt{48}}} g) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-frac{1}{sqrt{4}-sqrt{5}}+...+frac{1}{sqrt{15}-sqrt{16}}

Đọc tiếp

❤ Tính:

a) \(\sqrt{5-\sqrt{21}}-\sqrt{5+\sqrt{21}}\)

b)\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

c)\(\sqrt{7+\sqrt{24}}+\sqrt{31-\sqrt{600}}\)

d)\(\sqrt{28-\sqrt{300}}+\sqrt{4-\sqrt{12}}\)

e)\(\sqrt{7-\sqrt{40}}-\sqrt{5-\sqrt{24}}-\sqrt{6-\sqrt{20}}\)

f)\(\sqrt{48-10\sqrt{7+\sqrt{48}}}\)

g) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{15}-\sqrt{16}}\)

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Lớp 9 Toán Violympic toán 9

Yêu các anh như ARMY yêu...

17 tháng 8 2019 lúc 15:33

Rút gọn biểu thức :

a) \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)

b) \(B=\frac{2+\sqrt{6}+\sqrt{10}+\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{8}+\sqrt{12}+\sqrt{20}}.\frac{\sqrt{2}-1}{3}\)

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Lớp 9 Toán Violympic toán 9

thi thu thuy khuat

25 tháng 11 2019 lúc 12:50

tinh a. sqrt{5}-sqrt{48}+5sqrt{27}-sqrt{45} b.left(sqrt{5}+sqrt{2}right)left(3sqrt{2}-1right) c.3sqrt{50}-2sqrt{75}-4frac{sqrt{54}}{sqrt{3}}-3sqrt{frac{1}{3}} d.sqrt{left(sqrt{3}-3right)^2}+sqrt{4-2sqrt{3}} e.frac{5sqrt{2}-2sqrt{5}}{sqrt{5}-sqrt{2}}+frac{6}{2-sqrt{10}}-frac{20}{sqrt{10}} f.frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}+frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}+1}{sqrt{5}-1}

Đọc tiếp

tinh

a. \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)

b.\(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)

c.\(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)

d.\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

e.\(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)

f.\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

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Lớp 9 Toán Violympic toán 9

Anh Minh

1 tháng 4 2020 lúc 9:31

3. a.sqrt{left(4-sqrt{17}right)^2} b.frac{2sqrt{3}}{2} c frac{sqrt{6}+sqrt{14}}{text{2√3+√28}} d.frac{x+1}{sqrt{x^2-1}} e.frac{x^2-5}{x+sqrt{5}} f.frac{2}{2-sqrt{3}} g.frac{sqrt{2}+1}{sqrt{2}-1} f.frac{xsqrt{x}-1}{sqrt{x}-1} i.frac{3}{sqrt{20}}+frac{1}{sqrt{60}}-2sqrt{frac{1}{15}} k.frac{3}{sqrt{5}-sqrt{2}}+frac{4}{sqrt{6}+sqrt{2}} i.(frac{1}{sqrt{5}-sqrt{3}}+frac{1}{sqrt{5}+sqrt{3}})sqrt{5} h.left(sqrt{20}-sqrt{45}+sqrt{5}right)sqrt{5} l.left(5sqrt{3}+3sqrt{5}right):sqrt{15} m.frac...

Đọc tiếp

3. a.\(\sqrt{\left(4-\sqrt{17}\right)^2}\)

b.\(\frac{2\sqrt{3}}{2}\)

c \(\frac{\sqrt{6}+\sqrt{14}}{\text{2√3+√28}}\)

d.\(\frac{x+1}{\sqrt{x^2-1}}\)

e.\(\frac{x^2-5}{x+\sqrt{5}}\)

f.\(\frac{2}{2-\sqrt{3}}\)

g.\(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)

f.\(\frac{x\sqrt{x}-1}{\sqrt{x}-1}\)

i.\(\frac{3}{\sqrt{20}}+\frac{1}{\sqrt{60}}-2\sqrt{\frac{1}{15}}\)

k.\(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)

i.(\(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\))\(\sqrt{5}\)

h.\(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right)\sqrt{5}\)

l.\(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)

m.\(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{\frac{4}{3}}\)

n.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)

d\(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)

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Lớp 9 Toán Violympic toán 9

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