Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

giúp mk vs!!

`6, sqrt(5+ 2 sqrt 6) = sqrt(3 + 2 sqrt 3 . sqrt 2 + 2) = sqrt 3 + sqrt 2`

`9, sqrt(11-2 sqrt 30) = sqrt(5  -2 . sqrt 5 . sqrt 6 + 6) = sqrt 6 - sqrt 5`

`13, sqrt(7-2 sqrt 10) = sqrt(5- 2sqrt10 + 2) = sqrt 5 - sqrt 2`

Chia `2 -> 3` câu r mình làm cho ha.

Bạn làm câu 6, 9,13 giúp mình nha

a: \(6\sqrt{\dfrac{x}{2}y}=\dfrac{6\sqrt{xy}}{\sqrt{2}}=\dfrac{6\sqrt{2xy}}{2}=3\sqrt{2xy}\)

\(\Leftrightarrow2\sqrt{u-1}-\sqrt{u-1}+3\sqrt{u-1}=4\)

\(\Leftrightarrow4\sqrt{u-1}=4\)

=>u-1=1

hay u=2

\(=\dfrac{1}{2}y+\dfrac{3}{4}\cdot\left|2y-1\right|-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)

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

\(=4\cdot\dfrac{5}{2}\cdot\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\cdot\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{\left|3x\right|}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}=6\sqrt{x}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{2\left(\sqrt{7}+1\right)}=\dfrac{\sqrt{7}}{2}\)

\(=\dfrac{3+2\sqrt{2}}{9-8}+\dfrac{3+\sqrt{3}}{9-3}=3+2\sqrt{2}+\dfrac{3+\sqrt{3}}{6}=\dfrac{6\left(3+2\sqrt{2}\right)+3+\sqrt{3}}{6}=\dfrac{18+12\sqrt{2}+3+\sqrt{3}}{6}=\dfrac{21+\sqrt{3}+12\sqrt{2}}{6}\)

`a)2\sqrt{3x}-4\sqrt{3x}+27-3sqrt{3x}`   `(x >= 0)`

`=(2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x})+27`

`=-5\sqrt{3x}+27`

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`b)3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28`   `(x >= 0)`

`=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28`

`=(3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x})+28`

`=14\sqrt{2x}+28`