Đáp án:
\(\begin{array}{l}
b.\left( {\sqrt x - 12} \right)\left( {\sqrt x + 3} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.x - 5\sqrt x - 6 = x + \sqrt x - 6\sqrt x - 6\\
= \sqrt x \left( {\sqrt x + 1} \right) - 6\left( {\sqrt x + 1} \right)\\
= \left( {\sqrt x + 1} \right)\left( {\sqrt x - 6} \right)\\
b.x - \sqrt x - 12 = x - 4\sqrt x + 3\sqrt x - 12\\
= \sqrt x \left( {\sqrt x - 4} \right) + 3\left( {\sqrt x - 12} \right)\\
= \left( {\sqrt x - 12} \right)\left( {\sqrt x + 3} \right)\\
c.2x - 7\sqrt x + 5 = 2x - 2\sqrt x - 5\sqrt x + 5\\
= 2\sqrt x \left( {\sqrt x - 1} \right) - 5\left( {\sqrt x - 1} \right)\\
= \left( {\sqrt x - 1} \right)\left( {2\sqrt x - 5} \right)\\
e.x\sqrt x + 1 = \left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)\\
h.x\sqrt x - {2^3} = \left( {\sqrt x - 2} \right)\left( {x + 2\sqrt x + 4} \right)
\end{array}\)
