K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

11 tháng 1 2019

\(\sqrt{14+\sqrt{16900}}-\sqrt{19+\sqrt{900}}+\sqrt{45+\sqrt{3025}}\)

\(=\sqrt{14+\sqrt{130^2}}-\sqrt{19+\sqrt{30^2}}+\sqrt{45+\sqrt{55^2}}\)

\(=\sqrt{14+130}-\sqrt{19+30}+\sqrt{45+55}\)

\(=\sqrt{144}-\sqrt{49}+\sqrt{100}\)

\(=\sqrt{12^2}-\sqrt{7^2}+\sqrt{10^2}\)

\(=12-7+10\)

\(=5+10\)

\(=15\)

11 tháng 1 2019

Mk nghĩ bạn cx lm đc mak cần J đăng :)

\(\sqrt{14+\sqrt{16900}}-\sqrt{19+\sqrt{900}}+\sqrt{45+\sqrt{3025}}\)

\(=\sqrt{14+130}-\sqrt{19+30}+\sqrt{45+55}=12-7+10=15\)

Mk đang rảnh nên thik lm mấy câu dễ dễ "___"

12 tháng 7 2021

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)

2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)

\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)

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

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

\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)

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

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

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

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

12 tháng 7 2021

cảm ơn bn nhiều 

28 tháng 2 2017

ta có:\(\sqrt{45+27\sqrt{2}}\) +\(\sqrt{45-27\sqrt{2}}\) =3(\(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\))

phân thức cần biến đổi trở thành:\(\dfrac{\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}\).3

=\(\dfrac{10+2\sqrt{7}}{6\sqrt{2}}=\dfrac{5+\sqrt{7}}{\sqrt{2}}=\dfrac{5\sqrt{2}+\sqrt{14}}{2}\)

lấy phân số vừa tìm được cộng cho \(\dfrac{\sqrt{14}}{2}\) ta được giá trị biểu thức cần tìm là \(\dfrac{5\sqrt{2}}{2}\)

16 tháng 11 2017

Tại sao kết quả lại bằng, có thể giải thích hộ mình ko\(\dfrac{10+2\sqrt{7}}{6\sqrt{2}}\)

14 tháng 8 2023

\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{2}}{2}\)

___________

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

__________

\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)

\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)

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

\(=\dfrac{2}{3}\)

a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)

b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)

Rút gọn biểu thức: 1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\) 2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\) 3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\) 4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\) 5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\) 6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\) 7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\) 8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\) 9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\) 10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\) 11)...
Đọc tiếp

Rút gọn biểu thức:

1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\)

2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)

3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)

4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)

5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)

6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)

7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)

8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\)

9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\)

10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\)

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

12) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)

13) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

14) \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

15) \(\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{120}\)

16) \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)

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

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

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

20) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)

4
3 tháng 1 2019

1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)

2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)

3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2} \)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)

4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)

5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)

6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)

7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)

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


4 tháng 1 2019
https://i.imgur.com/pmexRQv.jpg
9 tháng 8 2020

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

9 tháng 8 2020

Hải Ngọc bạn làm mỗi 1 vế mà lại làm vế dễ nhát nữa thì chịu rồi

NV
12 tháng 7 2021

\(\sqrt{12-6\sqrt{3}}=\sqrt{9-6\sqrt{3}+3}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}\)

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

\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)

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

\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)

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

12 tháng 7 2021

\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

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

\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

30 tháng 9 2023

\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)

__

\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)

__

\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)

3 tháng 10 2021

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)

bạn giải chi tiết giúp mk đc k ạ

 

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

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

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)

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

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)

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

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

\(=\sqrt{7}-2+4-\sqrt{7}+8\)

=10