Bài 49 trang 29 SGK Toán 9 tập 1

Khử mẫu của biểu thức lấy căn

\(ab\sqrt{\dfrac{a}{b}};\,\,\, \dfrac{a}{b}\sqrt{\dfrac{b}{a}};\,\,\, \sqrt{\dfrac{1}{b}+\dfrac{1}{b^{2}}};\,\,\,\ \sqrt{\dfrac{9a^{3}}{36b}};\,\,\, 3xy\sqrt{\dfrac{2}{xy}}.\)

(Giả thiết các biểu thức có nghĩa).

+ Ta có

\(ab\sqrt{\dfrac{a}{b}}=ab\sqrt{\dfrac{a.b}{b.b}}=ab\sqrt{\dfrac{ab}{b^2}}=ab\dfrac{\sqrt{ab}}{\sqrt{b^2}}=ab\dfrac{\sqrt{ab}}{\left | b \right |}.\)

        Nếu \( b \ge 0\)  thì \(|b|=b \Rightarrow ab\dfrac{\sqrt{ab}}{\left | b \right |}=ab\dfrac{\sqrt{ab}}{b}=a\sqrt{ab}\). 

        Nếu \( b < 0\)  thì \(|b|=-b \Rightarrow ab\dfrac{\sqrt{ab}}{\left | b \right |}=-ab\dfrac{\sqrt{ab}}{b}=-a\sqrt{ab}\).

+ Ta có:

\( \dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\dfrac{a}{b}\sqrt{\dfrac{b.a}{a.a}}=\dfrac{a}{b}\sqrt{\dfrac{ab}{a^2}}\)

\(=\dfrac{a}{b}.\dfrac{\sqrt{ab}}{\sqrt{a^2}}\)\(=\dfrac{a}{b}.\dfrac{\sqrt{ab}}{|a|}\)\(=\dfrac{a\sqrt{ab}}{b|a|}\)

   Nếu \(a\geq 0\) thì \( |a|=a \Rightarrow \dfrac{a\sqrt{ab}}{b|a|}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b} .\)

   Nếu \(a<0\) thì  \(|a|=-a  \Rightarrow \dfrac{a\sqrt{ab}}{b|a|}=-\dfrac{a\sqrt{ab}}{ab}=-\dfrac{\sqrt{ab}}{b} .\)

+ Ta có:

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\sqrt{\dfrac{b}{b^2}+\dfrac{1}{b^2}}=\sqrt{\dfrac{b+1}{b^2}}\)

                    \(=\dfrac{\sqrt{b+1}}{\sqrt{b^2}}=\dfrac{\sqrt{b+1}}{|b|}\).

   Nếu \(b \ge 0\)  thì \(|b|=b \Rightarrow \dfrac{\sqrt{b+1}}{|b|}=\dfrac{\sqrt{b+1}}{b}\).

   Nếu \(-1 \le b < 0\)  thì \(|b|=-b \Rightarrow \dfrac{\sqrt{b+1}}{|b|}=-\dfrac{\sqrt{b+1}}{b}\).

+ Ta có:

\(\sqrt{\dfrac{9a^3}{36b}}=\sqrt{\dfrac{9}{36}}.\sqrt{\dfrac{a^3}{b}}=\sqrt{\dfrac{1}{4}}.\sqrt{\dfrac{a^3.b}{b.b}}\)

\(=\dfrac{1}{2}.\sqrt{\dfrac{a^2.ab}{b^2}}\)\(=\dfrac{1}{2}.\dfrac{\sqrt{a^2}.\sqrt{ab}}{\sqrt{b^2}}\)

\(=\dfrac{1}{2}.\dfrac{|a|\sqrt{ab}}{|b|}=\dfrac{|a|\sqrt{ab}}{2|b|}\).

Nếu \(a \ge 0,\ b \ge 0\) thì \(|a|=a,\ |b| =b \Rightarrow \dfrac{|a|\sqrt{ab}}{2|b|}=\dfrac{a\sqrt{ab}}{2b}\).

Nếu \(a < 0,\ b < 0\) thì \(|a|=-a,\ |b| =-b \Rightarrow \dfrac{|a|\sqrt{ab}}{2|b|}=\dfrac{a\sqrt{ab}}{2b}\).

+ Ta có:

\(3xy\sqrt{\dfrac{2}{xy}}=3xy.\sqrt{\dfrac{2.xy}{xy.xy}}=3xy.\dfrac{\sqrt{2xy}}{\sqrt{(xy)^2}}\)

\(=3xy.\dfrac{\sqrt{2xy}}{|xy|}\) \(=\dfrac{3xy.\sqrt{2xy}}{xy}=3\sqrt{2xy}\).


Lời giải


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