Lời giải:

\(=\sqrt{ab}.\sqrt{ab}+2.\sqrt{\frac{b}{a}}.\sqrt{ab}-\sqrt{\frac{a}{b}+\sqrt{\frac{1}{ab}}}.\sqrt{ab}\)

\(=ab+2\sqrt{b^2}-\sqrt{(\frac{a}{b}+\sqrt{\frac{1}{ab}}).ab}=ab+2|b|+\sqrt{a^2+\sqrt{ab}}\)

Điều kiện: \(\left\{{}\begin{matrix}b\ne0,m\ne0,n\ne0\\\dfrac{n}{m}\ge0\\mn\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b\ne0\\m\ne0\\n\ne0\\m,n\end{matrix}\right.\) ( 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

Èo. Câu này nhân vô đi bác. Nhân vô rồi rút gọn

Bài này quá đơn giản

\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)

chỗ đầu mình nhầm B = \(\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(....\right)\)

\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)

\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)

\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)

Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)

\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)

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 = \(\dfrac{1}{12}\)

Đăt\(\sqrt{a}\)=x, \(\sqrt{b}\)=y (x,y>0)
=>xy+1=4y => 4y≥ \(2\sqrt{xy}\)=>\(2\sqrt{y}\)≥\(\sqrt{x}\)=> 4y≥x=> 4≥ \(\dfrac{x}{y}\)=> \(\dfrac{1}{4}\)≤\(\dfrac{y}{x}\)=>\(\dfrac{-1}{4}\)≥\(\dfrac{-y}{x}\)
Xét:A=(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{y-x}\)+1):(\(\dfrac{xy+y}{x+y}\)+\(\dfrac{xy+x}{x-y}\)-1)
         = \(\dfrac{-2y^2\left(x+1\right)}{\left(x-y\right)\left(x+y\right)}\).\(\dfrac{\left(x-y\right)\left(x+y\right)}{2xy\left(x+1\right)}\)
=> A= \(\dfrac{-y}{x}\)≤\(\dfrac{-1}{4}\)
Dấu "=" xảy ra <=> xy=1 và x=4y <=> x=2, y=\(\dfrac{1}{2}\) <=> a =4, b=\(\dfrac{1}{4}\)

Vậy Max A =\(\dfrac{-1}{4}\) <=> a=4, b=\(\dfrac{1}{4}\)