Giúp mình với!mai nộp r!cảm ơn mn trc nhé!!!! (các bạn làm đầy đủ bước nhá) 32, \(\sqrt{15-6\sqrt{6}} \sqrt{33-12\sqrt{6}}\) 33, \(\sqrt{6-2\sqrt{5}} \sqrt{6 2\sqrt{5}}\) 34, \(\sqrt{8-2\sqrt{15}}...

2

LT

Lê Thị Thục Hiền

23 tháng 8 2019

32, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

=\(\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{33-2.3.2\sqrt{6}}\)

=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-2.3.2\sqrt{6}+9}\)

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

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

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

34, \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

=\(\sqrt{8-2.\sqrt{3}.\sqrt{5}}-\sqrt{23-2.2.\sqrt{5}.\sqrt{3}}\)

=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.\sqrt{3}+3}\)

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

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

LT

Lê Thị Thục Hiền

23 tháng 8 2019

35,\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

=\(\sqrt{16-2.4.\sqrt{15}+15}+\sqrt{15-2.3.\sqrt{15}+9}\)

=\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

=\(\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

=\(4-\sqrt{15}+\sqrt{15}-3\)

=1

36, \(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}\)

=\(\sqrt{49-2.5.\sqrt{24}}-\sqrt{49+2.5\sqrt{24}}=\sqrt{25-2.5.\sqrt{24}+24}-\sqrt{25+2.5.\sqrt{24}+24}=\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\)

=\(\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|=5-\sqrt{24}-5-\sqrt{24}=-2\sqrt{24}\)

37, \(\sqrt{3+2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

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

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...

YT

Yết Thiên

25 tháng 9 2021

1

TN

Tử Nguyệt Hàn

25 tháng 9 2021

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)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

BN

Bao Nguyen

1 tháng 10 2018

Giúp mình 2 câu này với ạ. Mình xin cảm ơn.

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

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

5

PM

Phùng Minh Quân

1 tháng 10 2018

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

\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)

\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)

\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)

\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)

\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)

\(=\)\(3\sqrt{6}+9\)

Chúc bạn học tốt ~

PM

Phùng Minh Quân

1 tháng 10 2018

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

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

\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) )

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

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

\(=\)\(1\)

Chúc bạn học tốt ~

PS : mới lớp 8 sai thì thông cảm >.<

M

manh

30 tháng 9 2023

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 : \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

2

H9

HT.Phong (9A5)

30 tháng 9 2023

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) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)

\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)

\(=\sqrt{6}-1+3-\sqrt{6}\)

\(=2\)

Đúng(2)

T

Toru

30 tháng 9 2023

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

\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)

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

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

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

\(=1\)

\(---\)

\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)

\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)

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

\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)

\(=2\sqrt{6}-3-5+2\sqrt{6}\)

\(=4\sqrt{6}-8\)

\(---\)

\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)

\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)

\(=\sqrt{6}-1+3-\sqrt{6}\)

\(=2\)

#\(Toru\)

Đúng(2)

3T

30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ

26 tháng 10 2023

Rút...

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...

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)

YT

Yết Thiên

25 tháng 9 2021

1

LL

Lấp La Lấp Lánh

25 tháng 9 2021

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

LD

La Đại Cương

14 tháng 7 2021

Cho mình hỏi:

a.\(\sqrt{15-6\sqrt{6}}+\sqrt{42-12\sqrt{6}}\)

b.1\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{24}+\sqrt{25}}\)

Trả lời giúp mình với ạ! mình cảm ơn!

1

DH

Dưa Hấu

14 tháng 7 2021

undefined

Đúng(1)

DT

Đào Thị Huyền

19 tháng 7 2018

rút gọn

C = \(\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}\)

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

F= \(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)

2

PK

Phùng Khánh Linh

19 tháng 7 2018

\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24+2.3.2\sqrt{6}+9}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\) \(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\) \(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)

TT

Trần Thiên Kim

19 tháng 7 2018

\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}+\sqrt{9}\right)^2}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\)

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

\(\Rightarrow\sqrt{2}D=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1-\sqrt{3}-1=-2\)

\(\Rightarrow D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)

\(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=27-2=25\)

Bài 1 Rút gọna) \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}\) b) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\) c) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\) Bài 2: Cho 2 đường thẳng (d): y = -x - 4 và (d₁): y = 3x + 2.a) Vẽ đồ thị (d) và (d₁) trên cùng một mặt phẳng tọa độ Oxy.b) Xác định tọa điểm A của 2 đường thẳng trên.c) Viết pt đường thẳng: (d₂): y = ax + b (a≠0) song song vs đường...

3T

30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ

8 tháng 12 2023

1

NL

Nguyễn Lê Phước Thịnh

8 tháng 12 2023

Câu 1:

a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)

\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)

\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)

\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

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

\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)

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

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

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

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

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

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

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

Bài 2:

a: loading...

b: Phương trình hoành độ giao điểm là:

\(3x+2=-x-4\)

=>4x=-6

=>x=-3/2

Thay x=-3/2 vào y=-x-4, ta được:

\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)

Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)

c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)

Vậy: (d2): y=-x+b

Thay x=-2 và y=5 vào (d2), ta được:

\(b-\left(-2\right)=5\)

=>b+2=5

=>b=5-2=3

Vậy: (d2): y=-x+3

LE

Limited Edition

28 tháng 9 2020

Rút gọn: \(\frac{\sqrt{12-2\sqrt{35}}+\sqrt{8-2\sqrt{15}}+\sqrt{5-2\sqrt{6}}}{\sqrt{12+2\sqrt{35}}-\sqrt{8+2\sqrt{15}}+\sqrt{5-2\sqrt{6}}}\)