Question

Giải phương trình

a) \(\sqrt{x^2-16+60}\) < x\(-\)6

b)\(\sqrt{x-3}\)\(-\)\(\sqrt{x-4}\) = 1

Answer 1

\(b)\sqrt {x - 3} - \sqrt {x - 4} = 1 \)

Điều kiện: \(x\ge4\)

\( \Leftrightarrow \sqrt {x - 3} = 1 + \sqrt {x - 4} \\ \Leftrightarrow x - 3 = 1 + 2\sqrt {x - 4} + x - 4\\ \Leftrightarrow - 3 = 3 + 2\sqrt {x - 4} \\ \Leftrightarrow 2\sqrt {x - 4} = 0\\ \Leftrightarrow \sqrt {x - 4} = 0\\ \Leftrightarrow x - 4 = 0\\ \Leftrightarrow x = 4\left( {TM} \right) \)

Vậy \(x=4\)

Answer 2

\(ĐK:\left[{}\begin{matrix}x\ge6\\x\le4\end{matrix}\right.\)\(\sqrt{x^2-16x+60}\ge0\Rightarrow\sqrt{x^2-16x+60}< x-6\Leftrightarrow x^2-16x+60< x^2-12x+36\Leftrightarrow-4x+24< 0\Leftrightarrow-4x< -24\Leftrightarrow x< 6\)\(\Rightarrow x\le4\)

\(b,ĐK:x\ge4;\sqrt{x-3}-\sqrt{x-4}=1\Leftrightarrow\sqrt{x-3}=\sqrt{x-4}+1\Leftrightarrow x-3=x-4+1+2\sqrt{x-4}\Leftrightarrow x-3=x-3+2\sqrt{x-4}\Leftrightarrow2\sqrt{x-4}=0\Leftrightarrow\sqrt{x-4}=0\Leftrightarrow x-4=0\Leftrightarrow x=4\left(\text{thoaman đkxđ}\right)\)