Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
a,\\
5\sqrt {25{a^2}} - 25a = 5\sqrt {{{\left( {5a} \right)}^2}} - 25a = 5.\left| {5a} \right| - 25a\\
= 5.\left( { - 5a} \right) - 25a = - 50a\,\,\,\,\,\,\,\,\,\,\,\left( {a < 0 \Rightarrow \left| a \right| = - a} \right)\\
b,\\
\sqrt {49{a^2}} + 3a = \sqrt {{{\left( {7a} \right)}^2}} + 3a = \left| {7a} \right| + 3a = 7a + 3a = 10a\,\,\,\,\,\,\,\,\,\,\left( {a \ge 0 \Rightarrow \left| a \right| = a} \right)\\
c,\\
\sqrt {16{a^4}} + 6{a^2} = \sqrt {{{\left( {4{a^2}} \right)}^2}} + 6{a^2} = \left| {4{a^2}} \right| + 6{a^2} = 4{a^2} + 6{a^2} = 10{a^2}\\
d,\\
A = 3\sqrt {9{a^6}} - 6{a^3} = 3.\sqrt {\left( {3{a^3}} \right)} - 6{a^3} = 3.\left| {3{a^3}} \right| - 6{a^3}\\
TH1:\,\,\,a \ge 0 \Rightarrow 3{a^3} \ge 0 \Rightarrow \left| {3{a^3}} \right| = 3{a^3}\\
\Rightarrow A = 3.3{a^3} - 6{a^3} = 3{a^3}\\
TH2:\,\,\,\,\,a < 0 \Rightarrow 3{a^3} < 0 \Rightarrow \left| {3{a^3}} \right| = - 3{a^3}\\
\Rightarrow A = 3.\left( { - 3{a^3}} \right) - 6{a^3} = - 15{a^3}\\
6,\\
a,\\
4x - \sqrt {{x^2} - 4x + 4} = 4x - \sqrt {{{\left( {x - 2} \right)}^2}} = 4x - \left| {x - 2} \right| = 4x - \left( {x - 2} \right) = 3x + 2\\
\left( {x \ge 2 \Rightarrow x - 2 \ge 0 \Rightarrow \left| {x - 2} \right| = x - 2} \right)\\
b,\\
3x + \sqrt {9 + 6x + {x^2}} = 3x + \sqrt {{{\left( {3 + x} \right)}^2}} = 3x + \left| {x + 3} \right| = 3x - \left( {x + 3} \right) = 2x - 3\\
\left( {x < - 3 \Rightarrow x + 3 < 0 \Rightarrow \left| {x + 3} \right| = - \left( {x + 3} \right)} \right)\\
c,\\
\dfrac{{x + 6\sqrt x + 9}}{{x - 9}} = \dfrac{{{{\left( {\sqrt x + 3} \right)}^2}}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} = \dfrac{{\sqrt x + 3}}{{\sqrt x - 3}}\\
d,\\
\dfrac{{\sqrt {{x^2} + 4x + 4} }}{{x + 2}} = \dfrac{{\sqrt {{{\left( {x + 2} \right)}^2}} }}{{x + 2}} = \dfrac{{\left| {x + 2} \right|}}{{x + 2}} = \pm 1
\end{array}\)
