I) Đk: x > 0 và x \(\ne\)9

\(D=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(D=\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(D=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

=> \(\frac{1}{D}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)

Để 1/D nguyên <=> \(\frac{2}{\sqrt{x}+1}\in Z\)

<=> \(2⋮\left(\sqrt{x}+1\right)\) <=> \(\sqrt{x}+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Do \(x>0\) => \(\sqrt{x}+1>1\) => \(\sqrt{x}+1=2\)

<=> \(\sqrt{x}=1\) <=> x = 1 (tm)

\(E=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\)

\(E=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(E=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Với x\(\ge\)0; ta có:

\(E=\frac{8}{9}\) <=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

<=> \(3\sqrt{x}=2x-2\sqrt{x}+2\)

<=> \(2x-4\sqrt{x}-\sqrt{x}+2=0\)

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

<=> \(\orbr{\begin{cases}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)

e) Ta có: \(E=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\ge0\forall x\in R\) (vì \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\))

Dấu "=" xảy ra<=> x = 0

Vậy MinE = 0 <=> x = 0

Lại có: \(\frac{1}{E}=\frac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}-1+\frac{1}{\sqrt{x}}\right)\ge\frac{3}{4}\left(2\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}-1\right)\)(bđt cosi)

=> \(\frac{1}{E}\ge\frac{3}{2}.\left(2-1\right)=\frac{3}{2}\)=> \(E\le\frac{2}{3}\)

Dấu "=" xảy ra<=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\) <=> x = 1

Vậy MaxE = 2/3 <=> x = 1

Edogawa Conan

câu 1 x không nguyên đâu bạn nhé, nên làm theo cách bạn là sai đấy

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

a) Thay x=4 zô là đc . ra kết quả \(\frac{7}{6}\)là dúng

b) \(B=\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)

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

\(=>P=A.B=\frac{3\sqrt{x}+1}{x+\sqrt{x}}.\frac{3\left(x+\sqrt{x}\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}=\frac{3}{3\sqrt{x}-1}\)

c) xét \(\frac{1}{P}=\frac{3\sqrt{x}-1}{3}\)

do \(\sqrt{x}\ge0=>3\sqrt{x}-1\ge-1\)\(=>\frac{3\sqrt{x}-1}{3}\ge-\frac{1}{3}\)

\(=>\frac{1}{P}\ge-\frac{1}{3}\)

dấu = xảy ra khi x=0

zậy ..

came ơn bạn nha!!!

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\(\in\left\{1;-1;3;-3\right\}\)

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

a/ \(P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x\ge0,x\ne4\right)\)

\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

b/ \(P=\frac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)

$P$ đạt giá trị lớn nhất \(\Leftrightarrow\left(\sqrt{x}+1\right)\) đạt GTNN

Vì \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\) đạt giá trị nhỏ nhất là $1$ tại \(x=0\)

Vậy \(MaxP=2\Leftrightarrow x=0\)

KL: ...................

a/ \(P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x\ge0,x\ne4\right)\)

\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

\(\Leftrightarrow P=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

b/ \(P=\frac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)

$P$ đạt giá trị lớn nhất \(\Leftrightarrow\left(\sqrt{x}+1\right)\) đạt GTNN

Vì \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\) đạt giá trị nhỏ nhất là $1$ tại \(x=0\)

Vậy \(MaxP=2\Leftrightarrow x=0\)

KL: ...................