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a. Ta có \(A=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}+\frac{9}{\sqrt{x}-3}\)
\(=3+\frac{9}{\sqrt{x}-3}\)
\(A\in Z\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\Rightarrow\sqrt{x}-3\in\left\{-9;-3;-1;1;3;9\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\Rightarrow x\in\left\{0;4;16;36;144\right\}\)
Vậy \(x\in\left\{0;4;16;36;144\right\}\)thì \(A\in Z\)
b. Thay \(x=7-4\sqrt{3}\Rightarrow A=\frac{3\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}-3}\)
\(=\frac{3\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-3}=\frac{3\left(2-\sqrt{3}\right)}{2-\sqrt{3}-3}=\frac{15-9\sqrt{3}}{2}\)
a: Khi x=6 thì \(A=\dfrac{4}{6-3}=\dfrac{4}{3}\)
b: \(B=\dfrac{4x}{x^2-9}-\dfrac{x-3}{x+3}\)(ĐKXĐ: \(x\notin\left\{3;-3\right\}\))
\(=\dfrac{4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}\)
\(=\dfrac{4x-\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{4x-x^2+6x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{-x^2+10x-9}{\left(x+3\right)\left(x-3\right)}\)
Bài 1:
Để B nguyên thì \(3x+1⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(4\right)\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
Bài 2:
a: Ta có: \(P=\dfrac{x^2-9}{x^2-6x+9}\)
\(=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{x+3}{x-3}\)
b: Để P nguyên thì \(x+3⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{4;2;5;1;6;0;9;-3\right\}\)
a) \(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\left[\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
c) để A>1/3
\(\Rightarrow\frac{\sqrt{x}+3-2}{\sqrt{x}+3}>\frac{1}{3}\)
\(\Rightarrow\frac{2}{\sqrt{x}+3}>\frac{2}{3}\)
\(\Rightarrow\sqrt{x}+3>3\)
\(\Rightarrow x>0\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
a: \(M=A+B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}-1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
\(=\dfrac{2x-6\sqrt{x}+x+2\sqrt{x}-3+11\sqrt{x}-3}{x-9}\)
\(=\dfrac{3x+7\sqrt{x}-6}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}-2\sqrt{x}-6}{x-9}=\dfrac{3\sqrt{x}-2}{\sqrt{x}-3}\)
b: M=M^4
=>M=0 hoặc M=1
=>3 căn x-2=căn x-3 hoặc 3 căn x-2=0
=>x=4/9