Điều kiện: \(\left\{{}\begin{matrix}be0,me0,ne0\\\dfrac{n}{m}\ge0\\mn\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}be0\\me0\e0\\m,n\end{matrix}\right.\) ( bổ sung chổ m,n nha chỗ đó là m,n cùng dấu )
Khi đó \(A=\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right).a^2.b^2\sqrt{\dfrac{n}{m}}\)
\(=\dfrac{am}{b}\sqrt{\dfrac{n}{m}}.a^2.b^2.\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}.a^2.b^2.\sqrt{\dfrac{n}{m}}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}.a^2.b^2.\sqrt{\dfrac{n}{m}}\)
\(=a^3bm\sqrt{\dfrac{n^2}{m^2}}-\dfrac{a^3b^3}{n}.\sqrt{\dfrac{mn^2}{m}}+a^4.\sqrt{\dfrac{mn}{nm}}\)
\(=a^3bm.\left|\dfrac{n}{m}\right|-\dfrac{a^3b^3}{n}.\sqrt{n^2}+a^4\)
\(=a^3bm.\dfrac{n}{m}-\dfrac{a^3b^3}{n}.\left|n\right|+a^4\) ( vì m,n cùng dấu )
\(=a^3bn-\dfrac{a^3b^3}{n}.\left|n\right|+a^4\)
Nếu n > 0 thì \(A=a^3bn-\dfrac{a^3b^3}{n}.n+a^4=a^3bn-a^3b^3+a^4\)
Nếu n < 0 thì \(A=a^3bn-\dfrac{a^3b^3}{n}.\left(-n\right)+a^4=a^3bn+a^3b^3+a^4\)
Vậy \(A=a^3bn-a^3b^3+a^4\) với n > 0; \(A=a^3bn+a^3b^3+a^4\) với n < 0
Bài này mới cơ bản thôi
