\(a,ĐK:x\ne-1;x\ne-2\\ b,A=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\dfrac{x-1}{x+2}\\ x=2020\Leftrightarrow A=\dfrac{2019}{2022}=\dfrac{673}{674}\\ c,A=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)

a,x^2- x # 0

hay x(x -1) # 0

<=> ĐKXĐ: x# 0

x -1 #0 => X# 1

bthay x= 0 vào A ta có

A= 2.0-1/ 0^2-0= -1

thay x= 3 vào A ta có

A= 2.3 -1/ 3^2-3 = 5/6

`a,`để `x` xác định thì

\(3x+9\ne0\)

\(\Leftrightarrow x\ne-3\)

`b,` tại `x=2` thì :

`A=(x^2 + 3)/(3x + 9) =(2^2 +3)/(3.2+9)=(4+3)/(6+9)=7/15`

`=>A=7/15`

\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)

b: \(=7\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(7-x+y\right)=0\)

ĐKXĐ: \(x\ne-3\)

Tại \(x=2\Rightarrow A=\dfrac{2^2+3}{3.2+9}=\dfrac{7}{15}\)

a) A xác định khi \(3x+9\ne0\Leftrightarrow x\ne-3\).

b) Với \(x=2\) thì \(A=\dfrac{2^2+3}{3\cdot2+9}=\dfrac{7}{15}\).

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

Khi x=-3 ta có:

\(\dfrac{\left(-3-1\right)^2}{-3+2}=\dfrac{\left(-4\right)^2}{-1}=-4\)

Khi x=1 ta có:
\(\dfrac{\left(1-1\right)^2}{1+2}=0\)

b) \(\dfrac{xy+3y^2}{x+y}=\dfrac{y\left(x+3y\right)}{x+y}\)

Khi x=3 y=-1 ta có:

\(\dfrac{-1\cdot\left(3+3\cdot-1\right)}{3\cdot-1}=0\)

a,A =  x² - x + 5 ,khi x = 2

= 2² - 2 + 5

= 7.

a: Thay x=2 vào A, ta được:

\(A=2^2-2+5=4+5-2=7\)