Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)

\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)

Sao ko bảo chị giải đi, kkkk

Haizzzzz

Câu a)

Câu b) Đề bài sai.

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

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

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

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

`a)(x-1)^2-(x-2)(x+2)`

`=x^2-2x+1-(x^2-4)`

`=-2x+5`

`b)(2x+4)(8x-3)(4x+1)^2`

`=(16x^2-6x+32x-12)(16x^2+8x+1)`

`=(16x^2-26x-12)(16x^2+8x+1)`

`=256x^4+128x^3+16x^2-416x^3-208x^2-26x-192x^2-96x-12`

`=256x^4-288x^3-384x^2-122x-12`

`c)(a+2)^3-a(a-3)^2`

`=a^3+6a^2+12a+8-a(a^2-6a+9)`

`=a^3+6a^2+12a+8-a^3+6a^2-9a`

`=12a^2+3a+8`

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

a) Ta có: 

b) Ta có: 

`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}`

`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`

`d)D=\sqrt{2+sqrt3}+sqrt{14-5sqrt3}+sqrt2`

`=>sqrt2D=sqrt{4+2sqrt3}+sqrt{28-10sqrt3}+2`

`=>sqrt2D=sqrt{(sqrt3+1)^2}+sqrt{(5-sqrt3)^2}+2`

`=>sqrt2D=8`

`=>D=4sqrt2`

a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

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

b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

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

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

\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)

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

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

a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)

b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115