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Gerda
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Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\dfrac{{{x^2} – 13}}{{x + \sqrt {13} }} = \dfrac{{\left( {x – \sqrt {13} } \right)\left( {x + \sqrt {13} } \right)}}{{x + \sqrt {13} }} = x – \sqrt {13} \\
b,\\
\dfrac{{{x^2} + 2\sqrt 3 .x + 3}}{{{x^2} – 3}} = \dfrac{{{{\left( {x + \sqrt 3 } \right)}^2}}}{{\left( {x – \sqrt 3 } \right)\left( {x + \sqrt 3 } \right)}} = \dfrac{{x + \sqrt 3 }}{{x – \sqrt 3 }}\\
c,\\
\sqrt {9{x^2}} – 2x = \sqrt {{{\left( {3x} \right)}^2}} – 2x = \left| {3x} \right| – 2x = – 3x – 2x = – 5x\\
d,\\
\sqrt {4 – 2\sqrt 3 } – \sqrt 3 = \sqrt {3 – 2.\sqrt 3 .1 + 1} – \sqrt 3 = \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} – \sqrt 3 = \left( {\sqrt 3 – 1} \right) – \sqrt 3 = – 1\\
e,\\
x – 4 + \sqrt {16 – 8x + {x^2}} = x – 4 + \sqrt {{x^2} – 2.x.4 + {4^2}} = \left( {x – 4} \right) + \sqrt {{{\left( {x – 4} \right)}^2}} = \left( {x – 4} \right) + \left| {x – 4} \right| = \left( {x – 4} \right) + \left( {x – 4} \right) = 2.\left( {x – 4} \right)\\
f,\\
\sqrt {11 + 6\sqrt 2 } – 3 + \sqrt 2 = \sqrt {9 + 2.3.\sqrt 2 + 2} – 3 + \sqrt 2 = \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} – 3 + \sqrt 2 = \left( {3 + \sqrt 2 } \right) – 3 + \sqrt 2 = 2\sqrt 2 \\
2,\\
a,\\
\sqrt {x + 4} = 2\,\,\,\,\,\,\,\,\,\left( {x \ge – 4} \right)\\
\Leftrightarrow x + 4 = 4\\
\Leftrightarrow x = 0\\
b,\\
\sqrt {2x – 1} = \sqrt 7 \,\,\,\,\,\,\,\left( {x \ge \dfrac{1}{2}} \right)\\
\Leftrightarrow 2x – 1 = 7\\
\Leftrightarrow x = 4\\
c,\\
\sqrt {x – 10} = – 2\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 10} \right)\\
\sqrt {x – 10} \ge 0 \Rightarrow ptvn\\
d,\\
\sqrt {4 – 7x} = 13\,\,\,\,\,\,\,\,\left( {x \le \dfrac{4}{7}} \right)\\
\Leftrightarrow 4 – 7x = 169\\
\Leftrightarrow 7x = – 165\\
\Leftrightarrow x = – \dfrac{{165}}{7}\\
3,\\
a,\\
\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| = 3y\\
b,\\
\dfrac{{\sqrt {49{a^6}{b^4}} }}{{\sqrt {128{a^8}{b^4}} }} = \sqrt {\dfrac{{49{a^6}{b^4}}}{{128{a^8}{b^4}}}} = \sqrt {\dfrac{{{7^2}}}{{{2^7}.{a^2}}}} = \dfrac{1}{{\sqrt 2 }}\left| {\dfrac{7}{{{2^3}.a}}} \right| = \dfrac{7}{{8\sqrt 2 a}}\\
c,\\
\sqrt {\dfrac{{x – 2\sqrt x + 1}}{{x + 2\sqrt x + 1}}} = \sqrt {\dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)}}} = \left| {\dfrac{{\sqrt x – 1}}{{\sqrt x + 1}}} \right| = \dfrac{{\left| {\sqrt x – 1} \right|}}{{\sqrt x + 1}}
\end{array}\)