Rút gọn:a. \(\sqrt{7 4\sqrt{3}}\)b.\(\sqrt{13-4\sqrt{3}}\)c.\(\sqrt{5-2\sqrt{6}}\)d.\(\sqrt{3 2\sqrt{2} \sqrt{6-4\sqrt{2}}}\)e.\(2 \sqrt{17-4\sqrt{9 4\sqrt{5}}}\)f.\(\sqrt{3a}.\sqrt{12a}\)với a < 0g.\...

Rút gọn:a. \(\sqrt{7+4\sqrt{3}}\)b.\(\sqrt{13-4\sqrt{3}}\)c.\(\sqrt{5-2\sqrt{6}}\)d.\(\sqrt{3+2\sqrt{2}+\sqrt{6-4\sqrt{2}}}\)e.\(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)f.\(\sqrt{3a}.\sqrt{12a}\)với a < 0g.\(\sqrt{16a^2b^8}\)h.\(\sqrt{7a}.\sqrt{63a^3}\)với a > 0i.\(\sqrt{2,7a}.\sqrt{5a^3}.\sqrt{1,5a^2}\)vớ a >...

0

HN

Hương Nguyễn

14 tháng 7 2018

1

KT

Không Tên

14 tháng 7 2018

a) \(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)

b) \(\sqrt{13-4\sqrt{3}}=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+1}\)

\(=\sqrt{\left(2\sqrt{3}-1\right)^2}=2\sqrt{3}-1\)

c) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

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

\(=\sqrt{3+2\sqrt{2}+\sqrt{\left(2-\sqrt{2}\right)^2}}\)

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

\(=\sqrt{5+\sqrt{2}}\)

e) \(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)

\(=2+\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}\)

\(=2+\sqrt{17-4\left(\sqrt{5}+2\right)}\)

\(=2+\sqrt{9-4\sqrt{5}}\)

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

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

f) đề sai nhé:

\(\sqrt{3a}.\sqrt{12a}=\sqrt{36a^2}=6a\)\(\left(a\ge0\right)\)

g) \(\sqrt{16a^2b^8}=4b^4\left|a\right|\)

h) \(\sqrt{7a}.\sqrt{63a^3}=\sqrt{441.a^4}=21a^2\)

3T

30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ

26 tháng 10 2023

Rút...

1

NL

Nguyễn Lê Phước Thịnh

26 tháng 10 2023

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)

\(=3+\sqrt{5}+3-\sqrt{5}=6\)

c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

\(=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}\)

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)

\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)

Đúng(3)

D

DŨNG

9 tháng 5 2022

Rút gọn:

A=\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)

B=\(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

2

NN

Nguyễn Ngọc Huy Toàn

9 tháng 5 2022

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(B=\left|\sqrt{5}+2\right|+\left|\sqrt{5}-2\right|\)

\(B=\sqrt{5}+2+\sqrt{5}-2\)

\(B=2\sqrt{5}\)

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NN

Nguyễn Ngọc Huy Toàn

9 tháng 5 2022

\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)

\(A=-1\)

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SV

Super Vegeta

23 tháng 5 2023

Rút gọn:

A= \(\sqrt{6-2\sqrt{5}}\) C= \(\sqrt{19-8\sqrt{3}}\)

B = \(\sqrt{4-\sqrt{12}}\) D= \(\sqrt{5-2\sqrt{6}}\)

2

2611

23 tháng 5 2023

`A=\sqrt{6-2\sqrt{5}}`

`A=\sqrt{(\sqrt{5}-1)^2}`

`A=\sqrt{5}-1`

_________

`B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}`

`B=\sqrt{(\sqrt{3}-1)^2}`

`B=\sqrt{3}-1`

_________

`C=\sqrt{19-8\sqrt{3}}`

`C=\sqrt{(4-\sqrt{3})^2}`

`C=4-\sqrt{3}`

_________

`D=\sqrt{5-2\sqrt{6}}`

`D=\sqrt{(\sqrt{3}-\sqrt{2})^2}`

`D=\sqrt{3}-\sqrt{2}`

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Y

YangSu

23 tháng 5 2023

\(A=\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5}^2-2\sqrt{5}+1^2}=\sqrt{ \left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)

\(D=\sqrt{5-2\sqrt{6}}=\sqrt{5-2.\sqrt{2}.\sqrt{3}}=\sqrt{\sqrt{3}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{2^2}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

VL

vũ linh

18 tháng 6 2021

tính:

a,\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

b,\(\sqrt{7-4\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

c,\(\dfrac{x-49}{\sqrt{x}-7}\)

d,\(\sqrt{4+2\sqrt{3}}-\sqrt{13+4\sqrt{3}}\)

e,\(2+\sqrt{17-4\sqrt{9+4\sqrt{45}}}\)

1

Y

Yeutoanhoc

18 tháng 6 2021

`a)\sqrt{9-4sqrt5}-sqrt5`

`=sqrt{5-2.2sqrt5+4}-sqrt5`

`=sqrt{(sqrt5-2)^2}-sqrt5`

`=|\sqrt5-2|-sqrt5`

`=sqrt5-2-sqrt5=-2`

`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`

`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`

`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`

`=|2-sqrt3|+|sqrt3-1|`

`=2-sqrt3+sqrt3-1=1`

`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`

`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`

`=sqrtx+7`

`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`

`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`

`=sqrt3+1-2sqrt3-1=-sqrt3`

`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)

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VL

vũ linh

18 tháng 6 2021

phần e bỏ số 4 ở cuối đi :))

DM

Đỗ Minh Anh

12 tháng 7 2019

Rút gọn biểu thức sau:

A = \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

B = \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

C = \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

D = \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

E = \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

2

B

💋Bevis💋

12 tháng 7 2019

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)

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

\(=2\sqrt{3}\)

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)

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

\(=2\sqrt{2}\)

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)

\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)

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

\(=6\)

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)

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

\(=4\)

\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)

\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)

\(=1+\sqrt{5}-\sqrt{5}+1\)

\(=2\)

K

KAl(SO4)2·12H2O

12 tháng 7 2019

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(A=\sqrt{3}+2+2-\sqrt{3}\)

A = 2 + 2

A = 4

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(B=\sqrt{2}+3+3-\sqrt{2}\)

B = 3 + 3

B = 6

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(C=3+2\sqrt{2}+3-2\sqrt{2}\)

C = 3 + 3

C = 6

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(D=\sqrt{5}+2-\sqrt{5}+2\)

D = 2 + 2

D = 4

\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(E=\sqrt{5}+1-\sqrt{5}+1\)

E = 1 + 1

E = 2

B1: rút gọn:a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)b, \(\sqrt{11+6\sqrt[]{2}}-3+\sqrt{2}\) c, \(x-4+\sqrt{16-8x+x^2}\) với x > 4d, \(\dfrac{x^2-5}{x+\sqrt{5}}\) x khác \(-\sqrt{5}\)e, \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}\) x khác \(-\sqrt{2}\)g, \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)giúp em với ạ , em cảm...

HP

Hương Phùng

10 tháng 7 2021

3

KH

Kiêm Hùng

10 tháng 7 2021

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Đúng(3)

AT

An Thy

10 tháng 7 2021

a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}\)

\(=2\sqrt{2}\)

c) \(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|\)

\(=x-4+x-4\left(x>4\right)=2x-8\)

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

e) \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}=\dfrac{\left(x+\sqrt{2}\right)^2}{x+\sqrt{2}}=x+\sqrt{2}\)

g) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{1}{\sqrt{2}}\)

Đúng(1)

KD

Khùng Điên

13 tháng 5 2018

a) \(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

b) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

c) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

d) \(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

e) \(\sqrt{9+4\sqrt{5}}\)

f) \(\sqrt{23+8\sqrt{7}}\)

2

NN

Nguyễn Như Ý

13 tháng 5 2018

a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)

\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)

\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)

b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)

c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)

d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)

\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)

e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)

f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)

DN

Đỗ Ngọc Diệp

5 tháng 9 2018

e)

....

24 tháng 6 2021

Rút gọn các biểu thức sau:

a \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

b \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

c \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

d \(\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

1

LT

Lê Thị Thục Hiền

24 tháng 6 2021

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)

\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...

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Y

Yeutoanhoc

24 tháng 6 2021

Khiếp CTV kìa sợ quá ;-;