Đáp án:
\(\begin{array}{l}
a,\\
3y\\
b,\\
- \dfrac{{\sqrt 2 {b^2}}}{{4a}}\\
c,\\
\left[ \begin{array}{l}
C = \dfrac{{3 - \sqrt x }}{{\sqrt x + 3}}\,\,\,\,\,\,\,\,\left( {0 \le x \le 9} \right)\\
C = \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}}\,\,\,\,\,\,\,\,\left( {x > 9} \right)
\end{array} \right.\\
d,\\
\left[ \begin{array}{l}
D = 2\,\,\,\,\,\,\,\left( {1 \le x \le 2} \right)\\
D = 2\sqrt {x - 1} \,\,\,\,\,\,\left( {x > 2} \right)
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
y > 0 \Rightarrow \left| y \right| = y\\
\dfrac{{\sqrt {63{y^3}} }}{{\sqrt {7y} }} = \sqrt {\dfrac{{63{y^3}}}{{7y}}} = \sqrt {9{y^2}} = \sqrt {{{\left( {3y} \right)}^2}} = \left| {3y} \right| = 3\left| y \right| = 3y\\
b,\,\,\,a < 0 \Rightarrow \left| a \right| = - a\\
{b^2} > 0,\,\,\,\forall b \ne 0 \Rightarrow \left| {{b^2}} \right| = {b^2}\\
\dfrac{{\sqrt {16{a^4}{b^6}} }}{{\sqrt {128{a^6}{b^2}} }} = \sqrt {\dfrac{{16{a^4}{b^6}}}{{128{a^6}{b^2}}}} = \sqrt {\dfrac{{{b^4}}}{{8{a^2}}}} = \dfrac{{\sqrt {{b^4}} }}{{\sqrt {8{a^2}} }} = \dfrac{{\sqrt {{{\left( {{b^2}} \right)}^2}} }}{{\sqrt {{{\left( {2\sqrt 2 a} \right)}^2}} }}\\
= \dfrac{{\left| {{b^2}} \right|}}{{\left| {2\sqrt 2 a} \right|}} = \dfrac{{{b^2}}}{{ - 2\sqrt 2 a}} = - \dfrac{{\sqrt 2 {b^2}}}{{4a}}\\
c,\\
C = \sqrt {\dfrac{{x - 6\sqrt x + 9}}{{x + 6\sqrt x + 9}}} = \sqrt {\dfrac{{{{\sqrt x }^2} - 2\sqrt x .3 + {3^2}}}{{{{\sqrt x }^2} + 2.\sqrt x .3 + {3^2}}}} \\
= \sqrt {\dfrac{{{{\left( {\sqrt x - 3} \right)}^2}}}{{{{\left( {\sqrt x + 3} \right)}^2}}}} = \dfrac{{\sqrt {{{\left( {\sqrt x - 3} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt x + 3} \right)}^2}} }} = \dfrac{{\left| {\sqrt x - 3} \right|}}{{\left| {\sqrt x + 3} \right|}} = \dfrac{{\left| {\sqrt x - 3} \right|}}{{\sqrt x + 3}}\\
TH1:\,\,\,0 \le x \le 9 \Leftrightarrow 0 \le \sqrt x \le 3\\
\Rightarrow \sqrt x - 3 \le 0 \Rightarrow \left| {\sqrt x - 3} \right| = - \left( {\sqrt x - 3} \right)\\
\Rightarrow C = \dfrac{{ - \left( {\sqrt x - 3} \right)}}{{\sqrt x + 3}} = \dfrac{{3 - \sqrt x }}{{\sqrt x + 3}}\\
TH2:\,\,\,x > 9 \Rightarrow \sqrt x > 3\\
\Rightarrow \sqrt x - 3 > 0 \Rightarrow \left| {\sqrt x - 3} \right| = \sqrt x - 3\\
\Rightarrow C = \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}}\\
\Rightarrow \left[ \begin{array}{l}
C = \dfrac{{3 - \sqrt x }}{{\sqrt x + 3}}\,\,\,\,\,\,\,\,\left( {0 \le x \le 9} \right)\\
C = \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}}\,\,\,\,\,\,\,\,\left( {x > 9} \right)
\end{array} \right.\\
d,\\
D = \sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } \,\,\,\,\,\left( {x \ge 1} \right)\\
= \sqrt {\left( {x - 1} \right) + 2\sqrt {x - 1} + 1} + \sqrt {\left( {x - 1} \right) - 2\sqrt {x - 1} + 1} \\
= \sqrt {{{\sqrt {x - 1} }^2} + 2.\sqrt {x - 1} .1 + {1^2}} + \sqrt {{{\sqrt {x - 1} }^2} - 2.\sqrt {x - 1} .1 + {1^2}} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 1} + 1} \right| + \left| {\sqrt {x - 1} - 1} \right|\\
= \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right|\\
TH1:\,\,\,1 \le x \le 2 \Leftrightarrow 0 \le x - 1 \le 1 \Leftrightarrow 0 \le \sqrt {x - 1} \le 1\\
\Rightarrow \sqrt {x - 1} - 1 \le 0 \Rightarrow \left| {\sqrt {x - 1} - 1} \right| = - \left( {\sqrt {x - 1} - 1} \right)\\
\Rightarrow D = \sqrt {x - 1} + 1 - \left( {\sqrt {x - 1} - 1} \right) = 2\\
TH2:\,\,\,x > 2 \Rightarrow x - 1 > 1 \Rightarrow \sqrt {x - 1} > 1\\
\Rightarrow \sqrt {x - 1} - 1 > 0 \Rightarrow \left| {\sqrt {x - 1} - 1} \right| = \sqrt {x - 1} - 1\\
\Rightarrow D = \sqrt {x - 1} + 1 + \left( {\sqrt {x - 1} - 1} \right) = 2\sqrt {x - 1} \\
\Rightarrow \left[ \begin{array}{l}
D = 2\,\,\,\,\,\,\,\left( {1 \le x \le 2} \right)\\
D = 2\sqrt {x - 1} \,\,\,\,\,\,\left( {x > 2} \right)
\end{array} \right.
\end{array}\)
