1) \(\dfrac{5-x}{x^2-3x}=\dfrac{5-x}{x\left(x-3\right)}\left(đk:x\ne0,x\ne3\right)\)

2) \(\dfrac{3x}{2x+3}\left(đk:x\ne-\dfrac{3}{2}\right)\)

a) ĐKXĐ: 

\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

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

c) Thay x = 3 vào A ta có:

\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

a) ĐKXĐ: 

\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)

\(\Leftrightarrow3x\ne\pm y\) 

b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)

\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)

\(B=\dfrac{2\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}\)

\(B=\dfrac{2}{3x+y}\)

Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:

\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)

a.  \(x^2-5x\ne0\)

=> ĐKXĐ: \(x\left(x-5\right)\ne0\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)

b. \(\dfrac{x^2-10x+25}{x^2-5x}\)

= \(\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}\)

= \(\dfrac{x-5}{x}\)

\(x\ne-2\)

ĐKXĐ: \(x\ne0;y\ne0\)

\(ĐK:x\ne2y;x\ne0;y\ne0\)

`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`

`<=>x ne 4,x me -1`

`b,ĐKXĐ:4x^2-25 ne 0`

`<=>(2x-5)(2x+5) ne 0`

`<=>x ne +-5/2`

`c,ĐKXĐ:8x^3+27 ne 0`

`<=>8x^3 ne -27`

`<=>2x ne -3`

`<=>x ne -3/2`

`d,2x+2 ne 0,4y^2-9 ne 0`

`<=>2x ne -2,(2y-3)(2y+3) ne 0`

`<=>x ne -1,y ne +-3/2`

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(B=\dfrac{6-7x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{2}{2-x}\)

\(=\dfrac{6-7x+3x-6+2x+4}{\left(x+2\right)\left(x-2\right)}\)

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

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

a. \(x\ne5\) là ĐKXĐ của biểu thức P

b. P =\(\dfrac{\left(x-5\right)^2}{x-5}\)=\(x-5\)

c. P = -1 <=> x-5 =-1 <=> x=4

\(a,ĐK:x^2-1=\left(x-1\right)\left(x+1\right)\ne0\Leftrightarrow x\ne\pm1\\ \dfrac{3x+3}{x^2-1}=\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x-1}=2\\ \Leftrightarrow x-1=\dfrac{3}{2}\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\\ b,\dfrac{3}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\left(tm\right)\)