Ôn tập chương 1: Căn bậc hai. Căn bậc ba

a: \(P=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

b: Thay \(x=\dfrac{2}{2+\sqrt{3}}=4+2\sqrt{3}\) vào P,ta được:

\(P=\dfrac{\left(\sqrt{3}+1+1\right)^2}{\sqrt{3}+1}=\dfrac{7+4\sqrt{3}}{\sqrt{3}+1}\)

4:

a:căn 2<2

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

\(\sqrt{2+\sqrt{2+\sqrt{2}}}< 2\) 

...

=>a<2

Chứng minh tương tự, ta được b<3

2:

\(a^2=2+\sqrt{2+\sqrt{...}}\)

=>a^2=2+a

=>a^2-a-2=0

=>a=2

b^2=6+căn 6+căn(...)

=>b^2=b+6

=>b=3

\(=2+4\sqrt{5}+\sqrt{17-12}=2+5\sqrt{5}\)

`a)\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}`

`=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}`

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

`=|\sqrt{5}-1|-|\sqrt{5}+1|`

`=\sqrt{5}-1-\sqrt{5}-1=-2`

`sqrt (5 - 2 sqrt 6) + sqrt 2) . 1/sqrt3`

`= sqrt((sqrt3-sqrt2)^2 + sqrt 2) . 1/sqrt3`

`= sqrt 3 - sqrt 2 + sqrt 2 . 1/sqrt 3`

`= sqrt 3 . 1/sqrt 3`

`= 1`.

`a)\sqrt{x^2-2x+4}=x+2`         `ĐK: x >= -2`

`<=>x^2-2x+4=x^2+4x+4`

`<=>6x=0`

`<=>x=0` (t/m)

Vậy `S={0}`

`a)\sqrt{x^2+3}=x+1`         `ĐK: x >= -1`

`<=>x^2+3=x^2+2x+1`

`<=>2x=2`

`<=>x=1` (t/m)

Vậy `S={1}`

a: \(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{2}\)

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

b: \(=\sqrt{x}+1-\sqrt{x}+1-1=1\)