Đáp án:
\(\begin{array}{l}
B1:\\
a){\left( {3 - \sqrt 3 } \right)^2}\\
b){\left( {\sqrt 6 + \sqrt 5 } \right)^2}\\
c){\left( {3 - \sqrt 5 } \right)^2}\\
B2:\\
a)\dfrac{5}{x}\\
b)\dfrac{{4b}}{5}\\
c) - \dfrac{1}{{3x}}\\
B3:\\
a)4a - 4\\
b)x\left( {x + 3} \right)\\
c){y^2} - 2y
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)12 - 2.3\sqrt 3 \\
= 9 - 2.3.\sqrt 3 + 3\\
= {\left( {3 - \sqrt 3 } \right)^2}\\
b)11 + 2\sqrt {30} = 6 + 2.\sqrt 6 .\sqrt 5 + 5\\
= {\left( {\sqrt 6 + \sqrt 5 } \right)^2}\\
c)14 - 2.3\sqrt 5 \\
= 9 - 2.3\sqrt 5 + 5\\
= {\left( {3 - \sqrt 5 } \right)^2}\\
B2:\\
a)\dfrac{{5\left| x \right|\sqrt {3x} }}{{{x^2}\sqrt {3x} }} = \dfrac{{5\left| x \right|}}{{{x^2}}}\\
= \dfrac{{5x}}{{{x^2}}} = \dfrac{5}{x}\left( {do:x > 0} \right)\\
b)\dfrac{{\sqrt {80a{b^2}} }}{{\sqrt {125a} }} = \dfrac{{\sqrt {16{b^2}} }}{{\sqrt {25} }}\\
= \dfrac{{4\left| b \right|}}{5} = \dfrac{{4b}}{5}\left( {do:b > 0} \right)\\
c)\sqrt {\dfrac{{81{x^4}{y^6}}}{{729{x^6}{y^6}}}} = \sqrt {\dfrac{1}{{9{x^2}}}} = \dfrac{1}{{3\left| x \right|}}\\
= - \dfrac{1}{{3x}}\left( {do:x < 0} \right)\\
B3:\\
a)\sqrt {16{{\left( {a - 1} \right)}^2}} = 4\left| {a - 1} \right| = 4\left( {a - 1} \right)\\
= 4a - 4\left( {do:a \ge 1} \right)\\
b)\sqrt {{x^2}{{\left( {x + 3} \right)}^2}} = \left| {x\left( {x + 3} \right)} \right|\\
= x\left( {x + 3} \right)\left( {do:x \ge 0} \right)\\
c)\sqrt {{y^2}{{\left( {y - 2} \right)}^2}} = \left| {y\left( {y - 2} \right)} \right|\\
= - y.\left( { - y + 2} \right)\left( {do:y < 0} \right)\\
= {y^2} - 2y
\end{array}\)
