\(a,\)
\(\sqrt{x-5}=3\)
\(\Leftrightarrow\)\(x-5=3^2\)
\(\Leftrightarrow\)\(x=14\)
\(b,\)
\(\sqrt{x-10}=-2\)
\(x\)không có giá trị ( vì \(\sqrt{x-10}\ge0\forall\))
\(c,\)
\(\sqrt{2x-1}=\sqrt{5}\)
\(\Leftrightarrow2x-1=\sqrt{5^2}\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
\(d,\)
\(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=12^2\)
\(\Leftrightarrow5x=4-144\)
\(\Leftrightarrow5x=-140\)
\(\Leftrightarrow x=-\frac{140}{5}=-28\)
a, \(\sqrt{x-5}=3;ĐK:x-5\ge0\Leftrightarrow x\ge5\)
Ta có: \(\sqrt{x-5}=3\Leftrightarrow x-5=9\Leftrightarrow x=14\)
b, \(\sqrt{x-10}=-2;ĐK:x-10\ge0\Leftrightarrow x\ge10\)
Vì: \(\sqrt{x-10}\ge0\) nên không có giá trị nào của x để \(\sqrt{x-10}=-2\)
c, \(\sqrt{2x-1}=\sqrt{5};ĐK:2x-1\ge0\Leftrightarrow x\ge0,5\)
Ta có: \(\sqrt{2x-1}=\sqrt{5}\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Leftrightarrow x=3\)
d, \(\sqrt{4-5x}=12;ĐK:4-5x\ge0\Leftrightarrow x\le\frac{4}{5}\)
Ta có: \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\)
\(\Leftrightarrow-5x=140\Leftrightarrow x=-28\)
