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

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

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

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

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

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

8 tháng 8 2023

kh đúng

11 tháng 7 2023

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=\(\dfrac{1}{4}\)

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

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

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

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

b: Thay a=16 và b=4 vào P, ta được:

\(P=\dfrac{4+2}{16-4\cdot2+4}\cdot\dfrac{4+16\cdot4+4\cdot4\cdot2+16}{16+4\cdot2+4}\)

\(=\dfrac{6}{12}\cdot\dfrac{116}{28}=\dfrac{29}{14}\)

8 tháng 2 2021

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 = \(\dfrac{1}{12}\)

2 tháng 10 2021

\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)

\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Do \(a\ge0\)

\(\Rightarrow a\in\left\{0;2;3\right\}\)

 

Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)

\(=\dfrac{2}{a-1}\)

Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)

hay \(a\in\left\{2;0;3\right\}\)

4 tháng 7 2023

a, \(VT=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b=VP\) đpcm

b,\(VT=1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}-\dfrac{a^2-a}{a-1}=1-\sqrt{a}+\sqrt{a}-a=1-a=VP\) đpcm

4 tháng 7 2023

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