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3 tháng 1 2022

b: Để A là số nguyên thì \(2x+2⋮x+3\)

\(\Leftrightarrow x+3\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{-4;-1;-5;1;-7\right\}\)

20 tháng 8 2023

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

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

20 tháng 8 2023

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

\(A=\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)

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

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

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

19 tháng 8 2023

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

\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{5}{\sqrt{x}+2}\)

\(=\dfrac{5\left(4\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)

17 tháng 12 2021

tất cả số nguyên x >0 là được

18 tháng 12 2021

Để B>0 thì x+2>0

hay x>-2

11 tháng 11 2018

\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)

\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)

\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)

\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)

\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)

\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)

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

=.= hok tốt!!

12 tháng 7 2018

a, \(\frac{x+2}{5}=\frac{1}{x-2}\Rightarrow\left(x+2\right)\left(x-2\right)=5\Rightarrow x^2-2x+2x-4=5\Rightarrow x^2=9\Rightarrow x=\pm3\)

b, \(\frac{3}{x-4}=\frac{x+4}{3}\Rightarrow\left(x+4\right)\left(x-4\right)=9\Rightarrow x^2-4x+4x-16=9\Rightarrow x^2=25\Rightarrow x=\pm5\)

c, \(\frac{x+2}{2}=\frac{1}{1-x}\Rightarrow\left(x+2\right)\left(1-x\right)=2\Rightarrow x-x^2+2-2x=2\Rightarrow-x^2-x=0\Rightarrow-x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

a: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}=\frac{2\left(x-2y\right)+x+2y+4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x-4y+x+2y+4}{\left(x-2y\right)\left(x+2y\right)}=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)

b: \(\frac{2}{x+2}+\frac{4}{x-2}+\frac{5x+2}{4-x^2}\)

\(=\frac{2\left(x-2\right)+4\left(x+2\right)-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x-4+4x+8-5x-2}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

c: \(\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{4y^2-x^2}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)+4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x\left(x+2y\right)}{\left(x+2y\right)\left(x-2y\right)}=\frac{2x}{x-2y}\)

d: \(\frac{3x^2-x}{x-1}+\frac{x+2}{1-x}+\frac{3-2x^2}{x-1}\)

\(=\frac{3x^2-x-x-2+3-2x^2}{x-1}=\frac{x^2-2x+1}{x-1}\)

\(=\frac{\left(x-1\right)^2}{x-1}=x-1\)