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

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

ĐKXĐ: \(x\ne2\) nên với x = 2 thì P không được xác định

\(Q=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)

\(=\dfrac{3\left(x+5\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)

\(=\dfrac{3x+15+x-3-2\left(x+3\right)}{x^2-9}=\dfrac{2x+6}{x^2-9}=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x-3}\)

Tại x = 2 thì \(Q=\dfrac{2}{2-3}=\dfrac{2}{-1}=-2\)

b) Để P < 0 tức \(\dfrac{1}{x-2}< 0\) mà tứ là 1 > 0

nên để P < 0 thì x - 2 < 0 \(\Leftrightarrow x< 2\)

Vậy x < 2 thì P < 0

c) Để Q nguyên tức \(\dfrac{2}{x-3}\) phải nguyên

mà \(\dfrac{2}{x-3}\) nguyên khi x - 3 \(\inƯ_{\left(2\right)}\)

hay x - 3 \(\in\left\{-2;-1;1;2\right\}\)

Lập bảng :

x - 3 -1 -2 1 2

x 2 1 4 5

Vậy x = \(\left\{1;2;4;5\right\}\) thì Q đạt giá trị nguyên

a) \(\dfrac{20x^3}{11y^2}.\dfrac{55y^5}{15x}=\dfrac{20.5.11.x.x^2.y^2.y^3}{11.3.5.x.y^2}=\dfrac{20x^2y^3}{3}\)

b) \(\dfrac{5x-2}{2xy}-\dfrac{7x-4}{2xy}=\dfrac{5x-2-7x+4}{2xy}=\dfrac{-2x+2}{2xy}=\dfrac{2\left(1-x\right)}{2xy}=\dfrac{1-x}{xy}\)

a) A= \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\)

\(ĐK:3x^2-7x+2\ne0\)

\(\Leftrightarrow\orbr{\begin{cases}x\ne\frac{1}{3}\\x\ne2\end{cases}\left(^∗\right)}\)

=> 3x² + 5x + 2 =0

<=> 3x² + 3x + 2x +2 = 0

<=> 3x .( x + 1 ) + 2 .( x + 1 ) =0

<=> (  x + 1 )(3x + 2 ) =0

<=> \(\orbr{\begin{cases}x+1=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{-2}{3}\left(t/m\left(^∗\right)\right)\end{cases}}}\)

Vậy x = -2/3 

b) \(B=\frac{2x^2+10x+12}{x^3-4x}=0\left(ĐK:x\ne0;x^2\ne4\Leftrightarrow x\ne0;x\ne\pm2\right)\)

<=> 2x²+ 10x + 12 = 0

<=> x² + 5x+ 6 =0

<=> ( x + 2 ) ( x + 3 ) =0\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=-3\left(t/m\right)\end{cases}}\) 

Vậy x = -3 

c)\(C=\frac{x^3+x^2-x-1}{x^3+2x-5}=0\)                         \(ĐK:x^3+2x-5\ne0\left(^∗\right)\)

<=> x³ + x² -x -1 =0

<=> ( x - 1 )(x² + 2x + 1 ) 

<=> ( x-1 ) (x+1)² = 0

<=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(t/m\left(^∗\right)\right)\\x=-1\left(t/m\left(^∗\right)\right)\end{cases}}}\)

Vậy x = { 1 ; -1 }

a) A = \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\) (ĐKXĐ: x khác 1/3, x khác 2)

<=> 3x^2 + 5x - 2 = 0

<=> (3x - 1)(x + 2) = 0

<=> 3x - 1 = 0 hoặc x + 2 = 0

<=> 3x = 1 hoặc x = -2

<=> x = 1/3 (ktm) hoặc x = -2 (tm)

=> x = -2

b) B = \(\frac{2x^2+10x+12}{x^3-4x}=0\) (ĐKXĐ: x khác 0, x khác +-2)

<=> \(\frac{2\left(x^2+5x+6\right)}{x\left(x^2-4\right)}=0\)

<=> \(\frac{2\left(x+2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{2\left(x+3\right)}{x\left(x-2\right)}=0\)

<=> 2(x + 3) = 0

<=> x + 3 = 0

<=> x = -3

c) C = \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) (ĐKXĐ: x khác x^3 + 2x - 5)

<=> \(\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}=0\)

<=> \(\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}=0\)

<=> \(\frac{\left(x+1\right)\left(x-1\right)\left(x+1\right)}{x^3+2x-5}=0\)

<=> (x + 1)(x - 1) = 0

<=> x + 1 = 0 hoặc x - 1 = 0

<=> x = -1 hoặc x = 1