Lời giải:

Áp dụng BĐT Cô-si:

$x+4\geq 4\sqrt{x}$

$\Rightarrow x-\sqrt{x}+7=(x+4)-\sqrt{x}+3\geq 4\sqrt{x}-\sqrt{x}+3$

$=3(\sqrt{x}+1)$

$\Rightarrow M=\frac{\sqrt{x}+1}{x-\sqrt{x}+7}\leq \frac{\sqrt{x}+1}{3(\sqrt{x}+1)}=\frac{1}{3}$
Vậy $M_{\max}=\frac{1}{3}$
Giá trị này xác định tại $x=4$

\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (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............

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: 

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

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

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

Vì \(\dfrac{9}{\sqrt{x}-1}\le\dfrac{9}{0-1}=-9\Leftrightarrow D\le2-9=-7\)

Vậy \(D_{max}=-7\Leftrightarrow x=0\)

a) \(A=\dfrac{3x}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)

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

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

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

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

\(\dfrac{1}{M}=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\)

\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}-\dfrac{\sqrt{x}}{27}=\dfrac{27\sqrt{x}+54-x-5\sqrt{x}}{27\left(\sqrt{x}+5\right)}\)\(=\dfrac{-x+22\sqrt{x}+54}{27\left(\sqrt{x}+5\right)}\)

\(\Rightarrow\sqrt{x}.27B+135B=-x+22\sqrt{x}+54\)

\(\Leftrightarrow x+\sqrt{x}\left(27B-22\right)+135B-54=0\) (1)

Coi PT (1) là phương trình bậc 2 ẩn \(\sqrt{x}\)

PT (1) có nghiệm không âm \(\Leftrightarrow\left\{{}\begin{matrix}\Delta=729B^2-1728B+700\ge0\\S=22-27B\ge0\\P=135B-54\ge0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2}{5}\le B\le\dfrac{14}{27}\)

Suy ra \(max_B=\dfrac{14}{27}\Leftrightarrow x=16\)

A làm tương tự