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Đặt \(a=\sqrt{x}>0\)
Khi đó:
\(A=\frac{a+3}{a^2+a+1}\)
\(\Rightarrow A\cdot a^2+A\cdot a+A=a+3\)
\(\Leftrightarrow A\cdot a^2+\left(A-1\right)\cdot a+\left(A-3\right)=0\)
Xét \(\Delta=\left(A-1\right)^2-4\left(A-3\right)A=A^2-2A+1-4A^2+12A\)
\(=-3A^2+10A+1\ge0\)
\(\Leftrightarrow\frac{5-2\sqrt{7}}{3}\le A\le\frac{5+2\sqrt{7}}{3}\)
Số xấu nên không chắc

a) \(A=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=-1+\frac{4}{\sqrt{x}-2}\)
Để \(A\in Z\Leftrightarrow\sqrt{x}-2\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-2;0;1;3;4;6\right\}\)
Mà \(x\in Z;\sqrt{x}\ge0\Rightarrow x\in\left\{0;1;9;16;36\right\}\)
b)\(A=\frac{4\sqrt{x}-2+3}{2\sqrt{x}-1}=2+\frac{3}{2\sqrt{x}-1}\)
Để \(A\in Z\Leftrightarrow2\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{-2;0;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-1;0;1;2\right\}\Leftrightarrow x\in\left\{0;1;4\right\}\)

#)Giải :
Bài 1 :
a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\frac{\left(1-x\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x+1}\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b) Để \(P>0\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow0< x< 1}\)
c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu ''='' xảy ra khi \(x=\frac{1}{4}\)

a) \(B=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{7\sqrt{x}-9}{x-9}\)
\(B=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x-9}-\frac{7\sqrt{x}-9}{x-9}\)
\(B=\frac{x+2\sqrt{x}-3-7\sqrt{x}+9}{x-9}\)
\(B=\frac{x-5\sqrt{x}+6}{x-9}\)
\(B=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
b) c) ?
b mình làm đc rồi, nó ko liên quan gì đến a và c đâu
Ta có : \(x-\sqrt{x}+1=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow\frac{2}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{2}{\frac{3}{4}}=\frac{8}{3}\)
hay : \(A\le\frac{8}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
Vậy : Max \(A=\frac{8}{3}\) tại \(x=\frac{1}{4}\)