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11 tháng 2 2018

Cần lắm không

11 tháng 2 2018

Có! Tết co giáo cho rõ nhiều bt :(

4 tháng 7 2021

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

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

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

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

b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)

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

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

c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)

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

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

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)

20 tháng 6 2021

`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`

`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

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

`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`

`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`

`=sqrt2(sqrt5+1+sqrt5-1)`

`=sqrt{2}.2sqrt5`

`=2sqrt{10}`

20 tháng 6 2021

`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`

`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`

`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`

`=2(7-3)`

`=8`

`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`

`=2/sqrt2=sqrt2`

17 tháng 7 2021

đó là số 2 ko phải chữ s mik xin lỗi

23 tháng 6 2019

a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)

\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{6-2\left(1+\sqrt{3}\right)}}\)

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

b) Tương tự a) đ/s =5

b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)

\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)

\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)

a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)

=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)

=>\(A=\sqrt{2\sqrt{2}+2}\)

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

=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)

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

=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)

=>\(B\simeq0,35\)

28 tháng 6 2021

`a)((sqrt(14)-sqrt7)/(1-sqrt2)+(sqrt{15}-sqrt5)/(1-sqrt3)):1/(sqrt7-sqrt5)`

`=((sqrt7(sqrt2-1))/(1-sqrt2)+(sqrt5(sqrt3-1))/(1-sqrt3)).(sqrt7-sqrt5)`

`=(-sqrt7-sqrt5)*(sqrt7-sqrt5)`

`=-(sqrt7+sqrt5)(sqrt7+sqrt5)`

`=-(7-5)=-2`

`b)sqrt2+1/sqrt{5+2sqrt6}+2/sqrt{8+2sqrt{15}}`

`=sqrt2+1/sqrt{3+2sqrt{3}.sqrt2+2}+2/sqrt{5+2sqrt{5}.sqrt3+3}`

`=sqrt2+1/sqrt{(sqrt3+sqrt2)^2}+2/sqrt{(sqrt5+sqrt3)^2}`

`=sqrt2+1/(sqrt3+sqrt2)+2/(sqrt5+sqrt3)`

`=sqrt2+((sqrt3+sqrt2)(sqrt3-sqrt2))/(sqrt3+sqrt2)+((sqrt5+sqrt3)(sqrt5-sqrt3))/(sqrt5+sqrt3)`

`=sqrt2+sqrt3-sqrt2+sqrt5-sqrt3=sqrt5`

a) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(-\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=-2\)

b) Ta có: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)

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

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

\(=\sqrt{5}\)

NV
18 tháng 6 2019

\(A=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\left(4-\sqrt{2}\right)}}}\)

\(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{6-2\sqrt{4+2\sqrt{3}}}}\)

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

\(=\sqrt{2}+\left(\sqrt{3}+1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

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

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

a) Ta có: \(2\sqrt{8}-3\sqrt{18}+\sqrt{32}\)

\(=4\sqrt{2}-6\sqrt{2}+4\sqrt{2}\)

\(=2\sqrt{2}\)

b) Ta có: \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(1+\sqrt{2}\right)^2}\)

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

\(=2\sqrt{2}\)

 

c) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

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

=1

 

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