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\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)
a: \(=ab+2\cdot\sqrt{\dfrac{b}{a}\cdot ab}-\sqrt{ab\cdot\left(\dfrac{a}{b}+\dfrac{1}{\sqrt{ab}}\right)}\)
\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)
\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)
b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)
\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)
\(=a^3nb-a^3b^3+a^4\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)
Lời giải:
\(=\sqrt{ab}.\sqrt{ab}+2.\sqrt{\frac{b}{a}}.\sqrt{ab}-\sqrt{\frac{a}{b}+\sqrt{\frac{1}{ab}}}.\sqrt{ab}\)
\(=ab+2\sqrt{b^2}-\sqrt{(\frac{a}{b}+\sqrt{\frac{1}{ab}}).ab}=ab+2|b|+\sqrt{a^2+\sqrt{ab}}\)