Đáp án:
\(\begin{array}{l}
2,\\
a,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
b,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\,\,\,\,\,b\\
4,\\
P = \dfrac{{ - 6}}{{\sqrt x + 3}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
DKXD:\,\,\,\,x \ge - 3\\
\dfrac{2}{3}\sqrt {9x + 27} - \dfrac{3}{2}\sqrt {4x + 12} - 2 = \sqrt {3 + x} \\
\Leftrightarrow \dfrac{2}{3}\sqrt {9\left( {x + 3} \right)} - \dfrac{3}{2}\sqrt {4.\left( {x + 3} \right)} - 2 = \sqrt {x + 3} \\
\Leftrightarrow \dfrac{2}{3}\sqrt {{3^2}.\left( {x + 3} \right)} - \dfrac{3}{2}\sqrt {{2^2}.\left( {x + 3} \right)} - 2 = \sqrt {x + 3} \\
\Leftrightarrow \dfrac{2}{3}.3\sqrt {x + 3} - \dfrac{3}{2}.2\sqrt {x + 3} - 2 = \sqrt {x + 3} \\
\Leftrightarrow 2\sqrt {x + 3} - 3\sqrt {x + 3} - 2 - \sqrt {x + 3} = 0\\
\Leftrightarrow - 2\sqrt {x + 3} - 2 = 0\\
\Leftrightarrow - 2\sqrt {x + 3} = 2\\
\Leftrightarrow \sqrt {x + 3} = - 1\\
\sqrt {x + 3} \ge 0 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
b,\\
DKXD:\,\,\,25{x^2} - 30x + 9 \ge 0\\
\sqrt {25{x^2} - 30x + 9} = x - 1\\
\Leftrightarrow \sqrt {{{\left( {5x} \right)}^2} - 2.5x.3 + {3^2}} = x - 1\\
\Leftrightarrow \sqrt {{{\left( {5x - 3} \right)}^2}} = x - 1\\
\Leftrightarrow \left| {5x - 3} \right| = x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
\left[ \begin{array}{l}
5x - 3 = x - 1\\
5x - 3 = - \left( {x - 1} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
5x - 3 - x + 1 = 0\\
5x - 3 + x - 1 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
4x - 2 = 0\\
6x - 4 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{2}{3}
\end{array} \right.
\end{array} \right.\\
\Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
a > b \ge 0 \Rightarrow \sqrt a > \sqrt b \Rightarrow \sqrt a .\sqrt b > {\sqrt b ^2} \Leftrightarrow \sqrt {ab} > b\\
\sqrt {ab} > b \Rightarrow \sqrt {ab} - b > 0\\
\dfrac{{a\sqrt b + b}}{{a - b}}.\sqrt {\dfrac{{ab + {b^2} - 2\sqrt {a{b^3}} }}{{a\left( {a + 2\sqrt b } \right) + b}}} :\dfrac{1}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{{{\sqrt a }^2} - {{\sqrt b }^2}}}.\sqrt {\dfrac{{ab + {b^2} - 2.\sqrt {{b^2}.ab} }}{{{a^2} + 2.a\sqrt b + b}}} :\dfrac{1}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\sqrt {\dfrac{{ab + {b^2} - 2.\left| b \right|.\sqrt {ab} }}{{{a^2} + 2a.\sqrt b + {{\sqrt b }^2}}}} :\dfrac{1}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\sqrt {\dfrac{{ab + {b^2} - 2.b.\sqrt {ab} }}{{{{\left( {a + \sqrt b } \right)}^2}}}} :\dfrac{1}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\sqrt {\dfrac{{{{\left( {\sqrt {ab} - b} \right)}^2}}}{{{{\left( {a + \sqrt b } \right)}^2}}}} :\dfrac{1}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\left| {\dfrac{{\sqrt {ab} - b}}{{a + \sqrt b }}} \right|:\dfrac{1}{{\sqrt a + \sqrt b }}\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\dfrac{{\sqrt {ab} - b}}{{a + \sqrt b }}.\left( {\sqrt a + \sqrt b } \right)\\
= \dfrac{{\sqrt b .\left( {a + \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}.\dfrac{{\sqrt b .\left( {\sqrt a - \sqrt b } \right)}}{{a + \sqrt b }}.\left( {\sqrt a + \sqrt b } \right)\\
= \sqrt b .\sqrt b \\
= b\\
4,\\
P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} + \dfrac{{3x + 3}}{{9 - x}}} \right).\left( {\dfrac{{\sqrt x - 7}}{{\sqrt x + 1}} + 1} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right).\left( {\dfrac{{\sqrt x - 7}}{{\sqrt x + 1}} + 1} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{{{\sqrt x }^2} - {3^2}}}} \right).\left( {\dfrac{{\sqrt x - 7}}{{\sqrt x + 1}} + 1} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right).\left( {\dfrac{{\sqrt x - 7}}{{\sqrt x + 1}} + 1} \right)\\
= \dfrac{{2\sqrt x .\left( {\sqrt x - 3} \right) + \sqrt x .\left( {\sqrt x + 3} \right) - \left( {3x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 7 + \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{2\sqrt x - 6}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{2.\left( {\sqrt x - 3} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3.\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{2\left( {\sqrt x - 3} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3.2}}{{\sqrt x + 3}}\\
= \dfrac{{ - 6}}{{\sqrt x + 3}}
\end{array}\)
