a, Ta có : \(\sqrt{x+1}>3\) đkxđ : \(x\ge-1\)
\(\Leftrightarrow x+1>3^2\)
\(\Leftrightarrow x+1>9\)
\(\Leftrightarrow x>8\)
Vậy bất phương trình có nghiệm là : x > 8
b,\(\sqrt{2x}< 1+\sqrt{3}\)
\(\Leftrightarrow2x< \left(1+\sqrt{3}\right)^2\)
\(\Leftrightarrow2x< 1+2\sqrt{3}+3\)
\(\Leftrightarrow2x< 4+2\sqrt{3}\)
\(\Leftrightarrow x< \frac{4+2\sqrt{3}}{2}\)
\(\Leftrightarrow x< 2+\sqrt{3}\)
c,\(x-4\sqrt{3}.\sqrt{x}+12=0\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2-2.2\sqrt{3}+\sqrt{x}+\left(2\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\sqrt{x}+2\sqrt{3}=0\)
\(\Leftrightarrow\sqrt{x}=-2\sqrt{3}\)
\(\Leftrightarrow x=\left(-2\sqrt{3}\right)^2\)
\(\Leftrightarrow x=12\)
ko bt có làm đúng ko nx ,hihi sai thì cho mk xin lỗi nha ko giúp đk bạn