Đáp án:
\(\begin{array}{l}
a,\\
\sqrt x \left( {\sqrt x - 1} \right)\\
b,\\
\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)\\
c,\\
\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)\\
d,\\
{\left( {\sqrt x + 1} \right)^2}\\
e,\\
\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)\\
f,\\
\left( {\sqrt x + 1} \right)\left( {2\sqrt x + 3} \right)\\
g,\\
\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\\
h,\\
\left( {\sqrt x - 1} \right)\left( {2\sqrt x - 3} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x - \sqrt x = {\sqrt x ^2} - \sqrt x = \sqrt x \left( {\sqrt x - 1} \right)\\
b,\\
x - 2\sqrt x - 3 = \left( {x - 3\sqrt x } \right) + \left( {\sqrt x - 3} \right)\\
= \left( {{{\sqrt x }^2} - 3\sqrt x } \right) + \left( {\sqrt x - 3} \right)\\
= \sqrt x \left( {\sqrt x - 3} \right) + \left( {\sqrt x - 3} \right)\\
= \left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)\\
c,\\
x - 16 = {\sqrt x ^2} - {4^2} = \left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)\\
d,\\
x + 2\sqrt x + 1 = {\sqrt x ^2} + 2.\sqrt x .1 + {1^2} = {\left( {\sqrt x + 1} \right)^2}\\
e,\\
x + \sqrt x - 6 = \left( {x - 2\sqrt x } \right) + \left( {3\sqrt x - 6} \right)\\
= \left( {{{\sqrt x }^2} - 2\sqrt x } \right) + \left( {3\sqrt x - 6} \right)\\
= \sqrt x \left( {\sqrt x - 2} \right) + 3.\left( {\sqrt x - 2} \right)\\
= \left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)\\
f,\\
2x + 5\sqrt x + 3 = \left( {2x + 2\sqrt x } \right) + \left( {3\sqrt x + 3} \right)\\
= \left( {2{{\sqrt x }^2} + 2\sqrt x } \right) + \left( {3\sqrt x + 3} \right)\\
= 2\sqrt x \left( {\sqrt x + 1} \right) + 3.\left( {\sqrt x + 1} \right)\\
= \left( {\sqrt x + 1} \right)\left( {2\sqrt x + 3} \right)\\
g,\\
2x + \sqrt x - 1 = \left( {2x - \sqrt x } \right) + \left( {2\sqrt x - 1} \right)\\
= \left( {2{{\sqrt x }^2} - \sqrt x } \right) + \left( {2\sqrt x - 1} \right)\\
= \sqrt x \left( {2\sqrt x - 1} \right) + \left( {2\sqrt x - 1} \right)\\
= \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\\
h,\\
2x - 5\sqrt x + 3 = \left( {2x - 2\sqrt x } \right) + \left( { - 3\sqrt x + 3} \right)\\
= \left( {2{{\sqrt x }^2} - 2\sqrt x } \right) - \left( {3\sqrt x - 3} \right)\\
= 2\sqrt x \left( {\sqrt x - 1} \right) - 3\left( {\sqrt x - 1} \right)\\
= \left( {\sqrt x - 1} \right)\left( {2\sqrt x - 3} \right)
\end{array}\)
