Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
x - 16 = {\sqrt x ^2} - {4^2} = \left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)\\
b,\\
x\sqrt x + 8 = {\sqrt x ^3} + {2^3} = \left( {\sqrt x + 2} \right).\left( {{{\sqrt x }^2} - \sqrt x .2 + {2^2}} \right) = \left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)\\
c,\\
x - 4 = {\sqrt x ^2} - {2^2} = \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)\\
d,\\
x\sqrt x - 1 = {\sqrt x ^3} - {1^3} = \left( {\sqrt x - 1} \right)\left( {x + \sqrt x .1 + {1^2}} \right) = \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)\\
2,\\
a,\\
x + 2\sqrt x = {\sqrt x ^2} + 2\sqrt x = \sqrt x \left( {\sqrt x + 2} \right)\\
b,\\
x - 3\sqrt x = {\sqrt x ^2} - 3.\sqrt x = \sqrt x .\left( {\sqrt x - 3} \right)\\
c,\\
x - \sqrt x - 2 = \left( {x - 2\sqrt x } \right) + \left( {\sqrt x - 2} \right) = \sqrt x .\left( {\sqrt x - 2} \right) + \left( {\sqrt x - 2} \right) = \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)\\
d,\\
x - 4\sqrt x + 3 = \left( {x - \sqrt x } \right) - \left( {3\sqrt x - 3} \right) = \sqrt x \left( {\sqrt x - 1} \right) - 3\left( {\sqrt x - 1} \right) = \left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)
\end{array}\)
