Đáp án:
\(\begin{array}{l}
e,\,\,\,\,21\sqrt {{x^2} + 1} \\
f,\,\,\,\,\sqrt 3 \\
g,\,\,\,\,3\sqrt {5a}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
e,\\
M = 4\sqrt {{x^2} + 1} - 2\sqrt {16\left( {{x^2} + 1} \right)} + 5\sqrt {25\left( {{x^2} + 1} \right)} \\
= 4\sqrt {{x^2} + 1} - 2\sqrt {16} .\sqrt {{x^2} + 1} + 5.\sqrt {25} .\sqrt {{x^2} + 1} \\
= 4\sqrt {{x^2} + 1} - 2.4.\sqrt {{x^2} + 1} + 5.5\sqrt {{x^2} + 1} \\
= 4\sqrt {{x^2} + 1} - 8\sqrt {{x^2} + 1} + 25\sqrt {{x^2} + 1} \\
= \left( {4 - 8 + 25} \right).\sqrt {{x^2} + 1} \\
= 21\sqrt {{x^2} + 1} \\
f,\\
x + y > 0 \Rightarrow \left| {x + y} \right| = x + y\\
N = \dfrac{2}{{x + y}}.\sqrt {\dfrac{{3{{\left( {x + y} \right)}^2}}}{4}} \\
= \dfrac{2}{{x + y}}.\sqrt {\dfrac{3}{4}} .\sqrt {{{\left( {x + y} \right)}^2}} \\
= \dfrac{2}{{x + y}}.\dfrac{{\sqrt 3 }}{2}.\left| {x + y} \right|\\
= \dfrac{2}{{x + y}}.\dfrac{{\sqrt 3 }}{2}.\left( {x + y} \right)\\
= \sqrt 3 \\
g,\\
a > \dfrac{1}{3} \Leftrightarrow 3a > 1 \Leftrightarrow 3a - 1 > 0 \Leftrightarrow 1 - 3a < 0\\
\Rightarrow \left| {1 - 3a} \right| = - \left( {1 - 3a} \right) = 3a - 1\\
P = \dfrac{3}{{3a - 1}}.\sqrt {5a.\left( {1 - 6a + 9{a^2}} \right)} \\
= \dfrac{3}{{3a - 1}}.\sqrt {5a} .\sqrt {1 - 6a + 9{a^2}} \,\,\,\left( {a > \dfrac{1}{3} \Rightarrow a > 0} \right)\\
= \dfrac{3}{{3a - 1}}.\sqrt {5a} .\sqrt {1 - 2.1.3a + {{\left( {3a} \right)}^2}} \\
= \dfrac{3}{{3a - 1}}.\sqrt {5a} .\sqrt {{{\left( {1 - 3a} \right)}^2}} \\
= \dfrac{3}{{3a - 1}}.\sqrt {5a} .\left| {1 - 3a} \right|\\
= \dfrac{3}{{3a - 1}}.\sqrt {5a} .\left( {3a - 1} \right)\\
= 3\sqrt {5a}
\end{array}\)
