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nhìn hơi đau mắt nhá bạn hoa mắt quá

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)

Ta có : \(P=\dfrac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{4}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x\left(x-3\right)+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x-x+6}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-3}{x-3}\)

b, Ta có : \(P=\dfrac{2x-3}{x-3}=\dfrac{2x-6+3}{x-3}=2+\dfrac{3}{x-3}\)

- Để P là số nguyên \(\Leftrightarrow x-3\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow x\in\left\{4;3;6;0\right\}\)

Vậy ...

a ĐKXĐ : \(x\ne2,x\ne3\)

\(\Rightarrow P=\dfrac{2x\left(x-3\right)+4-\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{x^2-5x+6}\)b Ta có P = \(\dfrac{2x^2-7x+6}{x^2-5x+6}=\dfrac{x^2-5x+6+x^2-2x}{x^2-5x+6}=1+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=1+\dfrac{x}{x-3}\)

Để P\(\in Z\) \(\Leftrightarrow1+\dfrac{x}{x-3}\in Z\) \(\Rightarrow\dfrac{x}{x-3}\in Z\) \(\Rightarrow x⋮x-3\) \(\Rightarrow x-3+3⋮x-3\)

\(\Rightarrow3⋮x-3\) \(\Rightarrow\left(x-3\right)\in\left\{-3;-1;1;3\right\}\) \(\Rightarrow x\in\left\{0;2;4;6\right\}\) 

Thử lại ta thấy đúng 

Vậy...

a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)

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

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)

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

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

a/

ĐKXĐ: \(x\ne\left\{-1;0;1\right\}\)

b.

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

c.

\(A=2\Rightarrow\dfrac{x+1}{x-1}=2\)

\(\Rightarrow x+1=2x-2\)

\(\Rightarrow x=3\) (thỏa mãn)

d.

\(A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}\)

\(A\) nguyên \(\Leftrightarrow\dfrac{2}{x-1}\) nguyên

\(\Rightarrow x-1=Ư\left(2\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-1=-2\\x-1=-1\\x-1=1\\x-1=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=0\left(ktm\right)\\x=2\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

Vậy \(x=\left\{2;3\right\}\) thì A nguyên