*Max
Xét `P-4`
`=(4\sqrtx+3-4x-4)/(x+1)`
`=(-4x+4\sqrtx-1)/(x+1)`
`=(-(2\sqrtx-1)^2)/(x+1)<=0`
`=>P<=1`
Dấu "=" `<=>2\sqrtx=1<=>x=1/4`
*Min
Xét `P+1`
`=(4\sqrtx+3+x+1)/(x+1)`
`=(x+4\sqrtx+4)/(x+1)`
`=(\sqrtx+2)^2/(x+1)>=0`
`=>P>=-1`
Dấu "=" `<=>\sqrtx+2=0<=>\sqrtx=-2`(vô lý)
=>Không có giá trị nhỏ nhất.

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

a) \(P=\dfrac{x^2-\sqrt[]{x}}{x+\sqrt[]{x}+1}-\dfrac{2x+\sqrt[]{x}}{\sqrt[]{x}}+\dfrac{2\left(x+\sqrt[]{x}-2\right)}{\sqrt[]{x}-1}\)

Điều kiện xác định \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left[\left(\sqrt[]{x}\right)^3-1\right]}{x+\sqrt[]{x}+1}-\dfrac{\sqrt[]{x}\left(2\sqrt[]{x}+1\right)}{\sqrt[]{x}}+\dfrac{2\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+2\right)}{\sqrt[]{x}-1}\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-1\right)\left(x+\sqrt[]{x}+1\right)}{x+\sqrt[]{x}+1}-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=\sqrt[]{x}\left(\sqrt[]{x}-1\right)-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=x-\sqrt[]{x}-2\sqrt[]{x}-1+2\sqrt[]{x}+4\)

\(\Rightarrow P=x-\sqrt[]{x}+3\)

b) \(A=\dfrac{P}{2012\sqrt[]{x}}=\dfrac{x-\sqrt[]{x}+3}{2012\sqrt[]{x}}\)\(\)

\(=\dfrac{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+3}{2012\sqrt[]{x}}\)

\(=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}{2012\sqrt[]{x}}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{\dfrac{11}{4}}{2012\sqrt[]{x}}=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\)

Ta lại có  \(\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}\ge0,\forall x\ne0\)

\(\dfrac{1}{\sqrt[]{x}}>0\Rightarrow\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{4.2012}=\dfrac{11}{8048}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{8048}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt[]{x}=1\Leftrightarrow x=1\)

Vậy \(GTNN\left(A\right)=\dfrac{11}{8048}\left(tạix=1\right)\)

\(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2.\left(\sqrt{x}+2\right)\)

\(=x-\sqrt{x}+3\)

b) \(\dfrac{P}{2012\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{2012\sqrt{x}}=\dfrac{\sqrt{x}}{2012}-\dfrac{1}{2012}+\dfrac{3}{2012\sqrt{x}}\)

\(=\left(\dfrac{\sqrt{x}}{2012}+\dfrac{3}{2012\sqrt{x}}\right)-\dfrac{1}{2012}\)

\(\ge2\sqrt{\dfrac{\sqrt{x}.3}{2012^2\sqrt{x}}}-\dfrac{1}{2012}\) (BĐT Cauchy)

\(=\dfrac{2\sqrt{3}}{2012}-\dfrac{1}{2012}=\dfrac{2\sqrt{3}-1}{2012}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{x}}{2012}=\dfrac{3}{2012\sqrt{x}}\Leftrightarrow x=3\)(tm)

sửa lại để:P=\(\dfrac{x}{\sqrt{x}-1}\)

a: \(P=\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\)

căn x-3>=-3

=>5/căn x-3<=-5/3

=>P<=-5/3+1=-2/3

Dấu = xảy ra khi x=0

\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))

\(=\left[\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)\(=\left[\dfrac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

\(=\left[\dfrac{2\sqrt{x}-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

\(=\dfrac{2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

\(=\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{2}{\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{M}{N}+1=\dfrac{2}{\sqrt{x}+2}+1\)

Ta thấy: \(\sqrt{x}\ge0\forall x\)

\(\Rightarrow\sqrt{x}+2\ge2\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\forall x\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+2}+1\le2\forall x\)

\(\Rightarrow Max_P=2\Leftrightarrow\dfrac{2}{\sqrt{x}+2}+1=2\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=1\)

\(\Leftrightarrow\sqrt{x}+2=2\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

#Urushi☕

Bạn tự rút gọn nha .

c) Ta có : \(P\text{=}\dfrac{M}{N}+1\text{=}\dfrac{2}{\sqrt{x}+2}+1\)

Để P có giá trị lớn nhất.

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}cóGTLN\)

\(\Leftrightarrow\sqrt{x}+2cóGTNN\)

Mà : \(\sqrt{x}+2\ge2\)

\(\Rightarrow\) Để : \(\left(\sqrt{x}+2\right)_{min}\) \(\Leftrightarrow\sqrt{x}\text{=}0\Leftrightarrow x\text{=}0\)

Vậy............

Sửa đề: Tìm giá trị lớn nhất

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(C=\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\)

\(\sqrt{x}-2>=-2\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{2}{\sqrt{x}-2}< =-1\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{2}{\sqrt{x}-2}+1< =-1+1=0\forall x\) thỏa mãn ĐKXĐ

=>\(C< =0\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Sửa đề: Tìm x để C đạt GTLN

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(C=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\)

\(\sqrt{x}-2>=-2\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{2}{\sqrt{x}-2}< =-\dfrac{2}{2}=-1\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{2}{\sqrt{x}-2}+1< =-1+1=0\forall x\) thỏa mãn ĐKXĐ

=>C<=0 với mọi x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Vậy: \(C_{max}=0\) khi x=0