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nguyenhiang thnag
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Nhóc vậy
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Sĩ Bí Ăn Võ
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Hải Nam Xiumin
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Lâm Vũ Thiên Phúc
1 tháng 7 2016 lúc 9:11

câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :

\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)

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

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

\(=3-\sqrt{6}+2\sqrt{6}-3\)   ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )

\(=\sqrt{6}\)

 

Mysterious Person
18 tháng 6 2017 lúc 10:20

A = \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

A = \(\dfrac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{5}+\sqrt{2}\right)}\)

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

C = \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

C = \(\left(4+\sqrt{15}\right).\left(\sqrt{40-10\sqrt{15}}-\sqrt{24-6\sqrt{15}}\right)\)

C = \(\left(4+\sqrt{15}\right)\left(\sqrt{\left(5-\sqrt{15}\right)^2}-\sqrt{\left(\sqrt{15}-3\right)^2}\right)\)

C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\left(\sqrt{15}-3\right)\right)\)

C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\sqrt{15}+3\right)\)

C = \(\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

C = \(32-8\sqrt{15}+8\sqrt{15}-30=2\)

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

D = \(\left(\sqrt{30-10\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\left(3+\sqrt{5}\right)\)

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

D = \(\left(5-\sqrt{5}-\left(\sqrt{5}-1\right)\right)\left(3+\sqrt{5}\right)\)

D = \(\left(5-\sqrt{5}-\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\)

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

D = \(18+6\sqrt{5}-6\sqrt{5}-10=8\)

E = \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{5}}\)

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

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

Trung123
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Võ Việt Hoàng
25 tháng 7 2023 lúc 7:30

\(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\dfrac{\sqrt{15-10\sqrt{2}}-\sqrt{10}}{\sqrt{5}}\)

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

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

\(=-3.-1=3\)

Vui lòng để tên hiển thị
25 tháng 7 2023 lúc 7:18

`((5-2 sqrt 5)/(2-sqrt5) - 2)((5+3 sqrt 5)/(3+sqrt 5) - 2)`

`= (5-2sqrt 5 - 4 + 2 sqrt 5)/(2 -sqrt 5) . (5+3 sqrt 5 - 6 - 2 sqrt 5)/(3 + sqrt 5)`

`= 1/(2-sqrt5) . (-1 + sqrt 5)/(3 + sqrt 5)`

`= (sqrt 5 - 1)/((sqrt 5 + 3)(2 - sqrt 5))`

`b, (sqrt(15-10sqrt2) - sqrt 10)/(sqrt 5)`

`= (sqrt(10 - 2 . sqrt 5.sqrt 10 + 5) - sqrt 10)/(sqrt 5)`

`= (sqrt 10 - sqrt 5 - sqrt 10)/(sqrt5)`

`= -1`

Alice dono
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Nguyễn Lê Phước Thịnh
18 tháng 7 2020 lúc 12:04

a) Ta có: \(\left(4+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{5}\right)\cdot\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

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

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

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

\(=\left(4+\sqrt{5}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{5}-2\sqrt{75}\)

b) Ta có: \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

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

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

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

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

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

\(=2\cdot\left(9-5\right)=2\cdot4=8\)

prayforme
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Nguyễn Lê Phước Thịnh
22 tháng 5 2022 lúc 13:07

a: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

b: \(=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

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

\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)

Nguyễn Ngọc Trình
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Nguyễn Tuấn Thành
4 tháng 8 2018 lúc 16:53

a.\(\Leftrightarrow\left(\sqrt{4+\sqrt{15}}\right)^2\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(\Leftrightarrow\sqrt{2}\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)

\(\Leftrightarrow\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4^2-15}\)

\(\Leftrightarrow\sqrt{5+2\sqrt{5.3}+3}\left(\sqrt{5}-\sqrt{3}\right)\)

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

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

Nguyễn Tuấn Thành
4 tháng 8 2018 lúc 17:29

b.T\(^2\)câu a ta tách (3+\(\sqrt{5}\))=\(\left(\sqrt{3+\sqrt{5}}\right)^2\);(\(\sqrt{10}-\sqrt{6}\))\(=\sqrt{2}\left(\sqrt{5}-1\right)\)

Rồi tính tiếp với đáp số =8

c.\(\Leftrightarrow\dfrac{\sqrt{2}\sqrt{\sqrt{5}+1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{\sqrt{2}\sqrt{\sqrt{5}+1}\sqrt{\sqrt{5}+1}}-\sqrt{2-2\sqrt{2}+1}\)

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

\(\Leftrightarrow\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}\left(\sqrt{5}+1\right)}-\left(\sqrt{2}-1\right)\)

\(\Leftrightarrow\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}\left(\sqrt{5}+1\right)}-\sqrt{2}+1\)

\(\Leftrightarrow\dfrac{3+\sqrt{5}+\sqrt{5}-1}{\sqrt{2}\left(\sqrt{5}+1\right)}-\sqrt{2}+1\)

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

Lê Thanh Ngọc
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