\(\sqrt{x-2}=4\)

ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=16\)

\(\Leftrightarrow\left|x-2\right|=16\)

\(\Leftrightarrow x-2=16\) (Vì \(x\ge2\))

\(\Leftrightarrow x=18\) (TM)

b) \(\sqrt{x-2}< 4\)

ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\left(\sqrt{x-2}\right)^2< 16\)

\(\Leftrightarrow\left|x-2\right|< 16\)

\(\Leftrightarrow x-2< 16\)

\(\Leftrightarrow x< 18\)

Kết hợp ĐK \(\Rightarrow2\le x< 18\)

\(a,9-4\sqrt{x}=1\Rightarrow-4\sqrt{x}=-8\)

\(\Rightarrow\sqrt{x}=2\Leftrightarrow x=4\)

\(b,\sqrt{\frac{x}{5}}=4\Rightarrow\frac{x}{5}=16\)

\(\Rightarrow x=5.16=80\)

\(c,\sqrt{7x}< 9\Leftrightarrow7x< 81\)

\(\Rightarrow x< \frac{81}{7}\)Và \(x\ge0\)

\(\Rightarrow0\le x< \frac{81}{7}\)

Đk:\(x\ge3;y\ge2021\)

\(A=x+y-\sqrt{x-3}.\sqrt{y-2021}\)

\(\Leftrightarrow A=\left(x-3\right)-\sqrt{x-3}.\sqrt{y-2021}+\dfrac{1}{4}\left(y-2021\right)+\dfrac{3}{4}\left(y-2021\right)+2024\)

\(\Leftrightarrow A=\left(\sqrt{x-3}-\dfrac{1}{2}\sqrt{y-2021}\right)^2+\dfrac{3}{4}\left(y-2021\right)+2024\ge2024\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y-2021=0\\\sqrt{x-3}-\dfrac{1}{2}\sqrt{y-2021}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=2021\\x=3\end{matrix}\right.\) (tm)

Vậy...

\(\sqrt{x}>2\Leftrightarrow x>4\)

\(5>\sqrt{x}\Leftrightarrow x< 25\)

\(\sqrt{x}< \sqrt{10}\Leftrightarrow x< 10\)( x không âm )

\(\sqrt{3x}< 3\Leftrightarrow3x< 9\Leftrightarrow x< 3\)

\(14\ge7\sqrt{2x}\Leftrightarrow\sqrt{2x}\le2\Leftrightarrow2x\le4\Leftrightarrow x\le2\)

Tham khảo nhé~

\(1.\sqrt{x}>2\left(Đk:x\ge0\right)\\ \Leftrightarrow\sqrt{x}>\sqrt{4}\)

\(\Leftrightarrow x>4\)

Đề thiếu rồi bạn

a) \(\sqrt{x}>4\) có nghĩa là \(\sqrt{x}>\sqrt{16}\)

Vì \(x\ge0\) (x không âm) nên \(\sqrt{x}>\sqrt{16}\Leftrightarrow x>16\)

Vậy \(x>16\)

b) \(\sqrt{4x}\le4\) có nghĩa là \(\sqrt{4x}\le\sqrt{16}\)

Vì \(x\ge0\) (x không âm) nên \(\sqrt{4x}\le\sqrt{16}\Leftrightarrow4x\le16\Leftrightarrow x\le4\)

Vậy \(x\le4\)

c) \(\sqrt{4-x}\ge6\) có nghĩa là \(\sqrt{4-x}\ge\sqrt{36}\)

Vì \(x\ge0\) (x không âm) nên \(\sqrt{4-x}\ge\sqrt{36}\Leftrightarrow4-x\ge36\Leftrightarrow x\le-32\)

Vậy \(x\le-32\)

\(\sqrt{3x}=9\)

\(\Rightarrow\left(\sqrt{3x}\right)^2=9^2\)

\(\Rightarrow3x=81\)

\(\Rightarrow x=27\)

\(\sqrt{3x}\)hay \(\sqrt{3}x\)vậy ?

cảm ơn mọi người