\(\sqrt{2x+27}-x=6\)

\(\Leftrightarrow2x+27=x^2+12x+36\)

\(\Leftrightarrow x^2+10x+9=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=-9\left(loại\right)\end{matrix}\right.\)

\(2\sqrt{3}\left(\sqrt{12}-\sqrt{27}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=2\sqrt{3}\left(\sqrt{3.2^2}-\sqrt{3.3^2}-3\sqrt{2}\right)+6\sqrt{6}\)

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

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

\(=-6\)

hết cứu đi mà làm

\(6\sqrt{6}-27=\left(\sqrt{6}-3\right)\left(9+3\sqrt{6}\right)\)

Giải PT?

\(\dfrac{\sqrt{27x}}{\sqrt{3}}=6\) (đk: \(x\ge0\))

\(\Leftrightarrow\sqrt{27x}=6\sqrt{3}\Leftrightarrow27x=108\Leftrightarrow x=4\) (tm)

Vậy pt có nghiệm duy nhất x=4

`sqrt{27x}/sqrt3=6`
`đk:x>=0`
`pt<=>sqrt{(27x)/3}=6`
`<=>sqrt{9x}=6`
`=>3sqrtx=6`
`<=>sqrtx=2`
`<=>x=4(tm)`
Vậy `S={4}`

\(a_1,\sqrt{x}< 7\\ \Rightarrow x< 49\\ a_2,\sqrt{2x}< 6\\ \Rightarrow x< 18\\ a_3,\sqrt{4x}\ge4\\ \Rightarrow4x\ge16\\ \Rightarrow x\ge4\\ a_4,\sqrt{x}< \sqrt{6}\\ \Rightarrow x< 6\)

\(b_1,\sqrt{x}>4\\ \Rightarrow x>16\\ b_2,\sqrt{2x}\le2\\ \Rightarrow2x\le4\\ \Rightarrow x\le2\\ b_3,\sqrt{3x}\le\sqrt{9}\\ \Rightarrow3x\le9\\ \Rightarrow x\le3\\ b_4,\sqrt{7x}\le\sqrt{35}\\ \Rightarrow7x\le35\\ \Rightarrow x\le5\)

Mình cám ơn Hà Quang Minh rất nhiều