ĐKXĐ: x>=0

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

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

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

\(\sqrt{x}+1>=1\forall x\) thỏa mãn ĐKXĐ

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

=>\(\dfrac{2}{\sqrt{x}+1}+1< =2+1=3\forall x\) thỏa mãn ĐKXĐ

=>P<=3 với mọi x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Đặt A=\(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+2}\)\(\Rightarrow Ax+A\sqrt{x}+2A-\sqrt{x}+1=0\)

\(\Leftrightarrow Ax+\sqrt{x}\left(A-1\right)+2A+1=0\)

\(\Delta=\left(A-1\right)^2-4A\left(2A+1\right)=A^2-2A+1-8A^2-4A\)\(=-7A^2-6A+1\ge0\)

\(\Rightarrow-1\le A\le\dfrac{1}{7}\)

Vậy Max A là \(\dfrac{1}{7}\)

Dâu"=" xảy ra \(\Leftrightarrow A=\dfrac{1}{7}\)

\(\Leftrightarrow7\sqrt{x}-7=x+\sqrt{x}+2\)

\(\Leftrightarrow x-6\sqrt{x}+9=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\Leftrightarrow x=9\)

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

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

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

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

Để \(P_{max}\) thì \(1+\dfrac{2}{\sqrt{x}-1}\) max

=>\(\dfrac{2}{\sqrt{x}-1}\) max

=>\(\sqrt{x}-1\) là số nguyên dương nhỏ nhất

=>\(\sqrt{x}-1=1\)

=>\(\sqrt{x}=2\)

=>x=4

Vậy: \(P_{max}=\dfrac{2+1}{2-1}=\dfrac{3}{1}=3\) khi x=4

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

1:

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

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

Xét \(2A=2\sqrt{x-2}+4\sqrt{x+1}+4038-2x\)     (Đk:\(x\ge2\))

     \(2A=-\left[\left(x-2\right)-2\sqrt{x-2}+1\right]-\left[\left(x+1\right)-4\sqrt{x+1}+2\right]+4042\)

   \(2A=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4042\le4042\)

\(\Leftrightarrow A\le2021\)

\(\Rightarrow Amax=2021\) khi x=3   (tm)Tự đăng câu hỏi xong tự trả lời (T-T)