\(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

P>3/2

=>P-3/2>0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

=>-căn x+2>0

=>-căn x>-2

=>0<x<4

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

\(ĐK:x\ge0;x\ne1;x\ne-2\)

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

\(P=\dfrac{3x+3\sqrt{x}-9+x-\sqrt{x}+3\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

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

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

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

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

Để P lớn nhất thì \(\sqrt{x}+2\) nhỏ nhất

Mà \(\sqrt{x}+2\ge2\) \(\Rightarrow Min=2\)

\(\Rightarrow P\le1+1+1+\dfrac{2}{2}=1+1+1+1=4\)

Vậy \(P_{max}=4\) khi \(x=0\)

a: \(P=\dfrac{\sqrt{x}+1-2\sqrt{x}+4+2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)

b: căn x+1>=1

=>P<=1

Dấu = xảy ra khi x=0

1) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

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

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

Để \(P=\dfrac{7}{2}\) thì \(2x+2\sqrt{x}+2-7\sqrt{x}=0\)

\(\Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{4}\end{matrix}\right.\)

a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

b: \(\sqrt{x}+3>=3\)

=>A<=1

Dấu = xảy ra khi x=0

c: \(P=A:\left(B-1\right)=\dfrac{3}{\sqrt{x}+3}:\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}-2}\)

Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{1;25\right\}\)

đkxđ: \(z\ge1;x\ge2;y\ge3\)

Đặt \(a=\sqrt{z-1}\ge0;b=\sqrt{x-2}\ge0;c=\sqrt{y-3}\ge0\)

\(\Rightarrow z=a^2+1;x=b^2+2;y=c^2+3\)

\(\Rightarrow A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+2}+\dfrac{c}{c^2+3}\)

Do các biến \(a,b,c\) độc lập nhau nên ta xét từng phân thức một.

Đặt \(f\left(a\right)=\dfrac{a}{a^2+1}\) \(\Rightarrow f\left(a\right).a^2-a+f\left(a\right)=0\) (*)

Nếu \(f\left(a\right)=0\) thì \(a=0\), rõ ràng đây không phải là GTLN cần tìm.

Xét \(f\left(a\right)\ne0\)

Để pt (*) có nghiệm thì \(\Delta=\left(-1\right)^2-4\left[f\left(a\right)\right]^2\ge0\) 

\(\Leftrightarrow\left(1+2f\left(a\right)\right)\left(1-2f\left(a\right)\right)\ge0\)

\(\Leftrightarrow-\dfrac{1}{2}\le f\left(a\right)\le\dfrac{1}{2}\)

\(f\left(a\right)=\dfrac{1}{2}\Leftrightarrow\dfrac{a}{a^2+1}=\dfrac{1}{2}\Leftrightarrow a^2+1=2a\Leftrightarrow a=1\) (nhận)

Vậy \(max_{f\left(a\right)}=\dfrac{1}{2}\).

 Tiếp đến, gọi \(g\left(b\right)=\dfrac{b}{b^2+2}\) \(\Rightarrow g\left(b\right).b^2-b+2g\left(b\right)=0\) (**)

 Tương tự nếu \(b=0\) thì vô lí. Xét \(b\ne0\). Khi đó để (**) có nghiệm thì \(\Delta=\left(-1\right)^2-8\left[g\left(b\right)\right]^2\ge0\)

\(\Leftrightarrow\left(1-2\sqrt{2}g\left(b\right)\right)\left(1+2\sqrt{2}g\left(b\right)\right)\ge0\)

\(\Leftrightarrow-\dfrac{1}{2\sqrt{2}}\le g\left(b\right)\le\dfrac{1}{2\sqrt{2}}\)

\(g\left(b\right)=\dfrac{1}{2\sqrt{2}}\Leftrightarrow\dfrac{b}{b^2+2}=\dfrac{1}{2\sqrt{2}}\Leftrightarrow b^2+2=2\sqrt{2}b\Leftrightarrow b=\sqrt{2}\) (nhận)

Vậy \(max_{g\left(b\right)}=\dfrac{1}{2\sqrt{2}}\)

Làm tương tự với \(h\left(c\right)=\dfrac{c}{c^2+3}\), ta được \(max_{h\left(c\right)}=\dfrac{1}{2\sqrt{3}}\), xảy ra khi \(c=\sqrt{3}\)

Vậy GTLN của A là \(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{6+3\sqrt{2}+2\sqrt{3}}{12}\), xảy ra khi \(\left(a,b,c\right)=\left(1,\sqrt{2},\sqrt{3}\right)\) hay \(\left(x,y,z\right)=\left(2,4,6\right)\).

Cái chỗ cuối mình sửa thế này nhé

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

P lớn nhất khi căn x+1=1

=>x=0

a) Ta có: \(B=\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)

\(=\left(\dfrac{x+3\sqrt{x}-3-\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)

\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\)