sao gặp toàn bài lp 9 z ????

\(=\frac{2\sqrt{5\cdot3}-2\sqrt{5\cdot2}+\sqrt{2\cdot3}-\sqrt{3\cdot3}}{2\sqrt{5}-2\sqrt{2\cdot5}-\sqrt{3}+\sqrt{2\cdot3}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(2\sqrt{5}-\sqrt{3}\right)}{-\left(\sqrt{2}-1\right)\cdot\left(2\sqrt{5}-\sqrt{3}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)

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

`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`

`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`

`<=>A^2=8+2sqrt{6-2sqrt5}`

`<=>A^2=8+2sqrt{(sqrt5-1)^2}`

`<=>A^2=8+2(sqrt5-1)`

`<=>A^2=6+2sqrt5=(sqrt5+1)^2`

`<=>A=sqrt5+1(do \ A>0)`

`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`

Vì `35+12sqrt6>35-12sqrt6`

`=>B>0`

`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`

`<=>B^2=70-2sqrt{361}`

`<=>B^2=70-2sqrt{19^2}=70-38=32`

`<=>B=sqrt{32}=4sqrt2(do \ B>0)`

`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`

`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`

`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`

`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`

`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`

`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`

`=(sqrt5+sqrt3)(sqrt5-sqrt3)`

`=5-3=2`

Biểu thức A chị tính A² rồi sẽ tính đc A

Biểu thức B ko bt có sai đề ở căn thứ 2 ko ạ

Nếu nhân B với căn 2 thì cái căn thức nhất tách đc thành hđt (a+b)² đấy ạ nhưng cái căn thứ 2 thì ko tách đc

đề câu B chả sai đi chỗ nào :)) tại tụi m tách sai thôi =)) 

\(B=\sqrt{29+6\sqrt{6}}-\sqrt{32-6\sqrt{15}}\)

\(B=\sqrt{\left(3\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}\) ( tách ra hằng đẳng thức ) 

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

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

nuột không :)) 

10)=5.382332347

11)=1.918230732

\(\dfrac{2\sqrt{15}-2\sqrt{10}-3+\sqrt{6}}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)

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

\(=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}=-\sqrt{3}-\sqrt{6}+\sqrt{2}+2\)

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

2.\(\sqrt{10-4\sqrt{6}}=\sqrt{4-4\sqrt{6}+6}=\sqrt{\left(2-\sqrt{6}\right)^2}=\left|2-\sqrt{6}\right|=\sqrt{6}-2\)       \(\sqrt{15-6\sqrt{6}}=\sqrt{9+6\sqrt{6}+6}=\sqrt{\left(3+\sqrt{6}\right)^2}=3+\sqrt{6}\)

=>\(\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)=\(3+\sqrt{6}-\sqrt{6}+2\)=5

3. Tương tự bằng :\(8-3\sqrt{6}\)

1) \(\sqrt{6+2\sqrt{5}}\) = \(\sqrt{1+2.1.\sqrt{5}+\sqrt{5}^2}\) =  \(\sqrt{\left(1+\sqrt{5}\right)^2}\)

2) \(\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

= \(\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6}^2}\) - \(\sqrt{2^2.2.2.\sqrt{6}+\sqrt{6}^2}\)

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

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

= 3 + \(\sqrt{6}\) - 2 + \(\sqrt{6}\)

= 1 + 2\(\sqrt{6}\)

3) \(\sqrt{31-10\sqrt{6}}-\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\sqrt{5^2-2.5.\sqrt{6}+\sqrt{6}^2}\) - \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)

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

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

= 5 - \(\sqrt{6}-3-2\sqrt{6}\)

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

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