HU HU MN ƠI GIÚP E VS E KO HIỂU CÁI J CẢ E ĐANG CẦN LẮM GIÚP ĐỠ E VS AK vote ngay 5* ak
1 .Rút gọn:
a) 2/ab³c² . √a²b⁴c⁶/4 với a<0 c>0
b) √75(a – 3)²/48(b + 1)⁴ với a>3 b≠1
c) (a + b)³/b² . √a³b/a + b . √(a + b)³/ab⁵ (a<0 b<0)
d) √a² -12a + 36/b³ . √b/(a² - 36)² (a>0 b<0 a≠0)
e) 2/a² - b² . √3(a + b)/4
f) √ab + 2√b/a - √a/b + √1/ab
Đáp án:
$\begin{array}{l}
a)\dfrac{2}{{a{b^3}{c^2}}}.\sqrt {\dfrac{{{a^2}{b^4}{c^6}}}{4}} \left( {a < 0;c > 0} \right)\\
= \dfrac{2}{{a{b^3}{c^2}}}.\dfrac{{ – a{b^2}{c^3}}}{2}\\
= – 1\\
b)\sqrt {\dfrac{{75{{\left( {a – 3} \right)}^2}}}{{48.{{\left( {b + 1} \right)}^4}}}} \\
= \sqrt {\dfrac{{25}}{{16}}} .\dfrac{{\left| {a – 3} \right|}}{{{{\left| {b + 1} \right|}^2}}}\\
= \dfrac{{5\left( {a – 3} \right)}}{{4{{\left( {b + 1} \right)}^2}}}\left( {do:a > 3;b \ne – 1} \right)\\
c)\dfrac{{{{\left( {a + b} \right)}^3}}}{{{b^2}}}.\sqrt {\dfrac{{{a^3}b}}{a}} + b.\sqrt {\dfrac{{{{\left( {a + b} \right)}^2}}}{{a{b^5}}}} \\
= \dfrac{{{{\left( {a + b} \right)}^3}}}{{{b^2}}}.\sqrt {{a^2}b} + b.\dfrac{{\left( {a + b} \right)}}{{{b^2}\sqrt {ab} }}\\
= \dfrac{{{{\left( {a + b} \right)}^3}}}{{b\sqrt b }}.a + \dfrac{{a + b}}{{b\sqrt {ab} }}\\
= \dfrac{{{{\left( {a + b} \right)}^2}.a\sqrt a + a + b}}{{b\sqrt {ab} }}\\
d)\sqrt {\dfrac{{{a^2} – 12a + 36}}{{{b^3}}}} .\sqrt {\dfrac{b}{{{{\left( {{a^2} – 36} \right)}^2}}}} \\
= \dfrac{{\sqrt {{{\left( {a – 6} \right)}^2}} }}{{b\sqrt b }}.\dfrac{{\sqrt b }}{{\left| {{a^2} – 36} \right|}}\\
= \dfrac{1}{{b.\left| {a + 6} \right|}}\\
= \dfrac{1}{{b.\left( {a + 6} \right)}}\\
e)\dfrac{2}{{{a^2} – {b^2}}}.\dfrac{{\sqrt 3 \left( {a + b} \right)}}{4}\\
= \dfrac{{\sqrt 3 }}{{2\left( {a – b} \right)}}\\
f)\sqrt {ab} + 2\sqrt {\dfrac{b}{a}} – \sqrt {\dfrac{a}{b}} + \sqrt {\dfrac{1}{{ab}}} \\
= \dfrac{{ab + 2b – a + 1}}{{\sqrt {ab} }}
\end{array}$