2√10.5√8.√2210.58.2

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√20.(5√3+√5)20.(53+5)

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12 tháng 8 2020

1)  Cách 1 :

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

\(M=\sqrt{9-6\sqrt{2}+2}+\sqrt{9+6\sqrt{2}+2}\)

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

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

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

Cách 2 :

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

\(\Rightarrow M^2=11-6\sqrt{2}+2\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}+11+6\sqrt{2}\)

\(\Leftrightarrow M^2=22+2.7=36\)

\(\Leftrightarrow M=6\left(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}>0\right)\)

2) 

\(A=53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}\)

\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{8-4\sqrt{2}+1}}\)

\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}\)

\(\Leftrightarrow A=53-20\sqrt{4+\left|2\sqrt{2}-1\right|}\)

\(\Leftrightarrow A=53-20\sqrt{4+2\sqrt{2}-1}\)

\(\Leftrightarrow A=53-20\sqrt{3+2\sqrt{2}}\)

\(\Leftrightarrow A=53-20\sqrt{2+2\sqrt{2}+1}\)

\(\Leftrightarrow A=53-20\left(\sqrt{2}+1\right)\)

\(\Leftrightarrow A=53-20\sqrt{2}-20=33-20\sqrt{2}\)

12 tháng 8 2020

3) 

\(M=\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(M=\sqrt{3-\sqrt{5}}.\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\)

\(M=\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)

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

\(\Rightarrow M^2=8.\left(3-\sqrt{5}\right).\left(5+2\sqrt{5}+1\right)\)

\(\Leftrightarrow M^2=\left(24-8\sqrt{5}\right)\left(6+2\sqrt{5}\right)\)

\(\Leftrightarrow M^2=144+48\sqrt{5}-48\sqrt{5}-80\)

\(\Leftrightarrow M^2=64\Leftrightarrow M=8\left(\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)>0\right)\)

a: \(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=\sqrt{\sqrt{3}}\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right)\)

\(=0\)

b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{8}-2\sqrt{5}-6\sqrt{5}\right)\)

\(=\sqrt{\sqrt{3}}\left(4\sqrt{2}-8\sqrt{5}\right)\)

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

a: \(=2\sqrt{\sqrt{3}}\cdot4\sqrt{5}-2\cdot\sqrt{\sqrt{3}}\cdot\sqrt{5}-3\cdot\sqrt{\sqrt{3}}\cdot2\sqrt{5}\)

\(=2\sqrt{\sqrt{3}}\left(4\sqrt{5}-\sqrt{5}-3\sqrt{5}\right)=0\)

b: \(=2\cdot2\sqrt{2}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\sqrt{5}\cdot\sqrt{\sqrt{3}}\)

\(=2\sqrt{\sqrt{3}}\left(2\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)=2\sqrt{\sqrt{3}}\cdot\left(2\sqrt{2}-4\sqrt{5}\right)\)

19 tháng 6 2018

e , \(\sqrt{11^2-\left(6\sqrt{2}\right)^2}\)

27 tháng 10 2019

g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath