a) x2-y2+3x+3y
=(x+y)(x-y)+3(x+y)
=(x+y)(x-y+3)
b) x2+2x+1-y2
=(x+1)2-y2
=(x-y+1)(x+y+1)
a) (x-2)2
= x2-2.x.2 + 22
=x2-4x+4
b) \(\frac{5x+3}{x+2}+\frac{7}{x+2}=\frac{5x+3+7}{x+2}=\frac{5x+10}{x+2}=\frac{5\left(x+2\right)}{x+2}=5\)
c) \(\frac{2a}{a-1}+\frac{a}{1-a}=\frac{2a}{a-1}-\frac{a}{a-1}=\frac{2a-a}{a-1}=\frac{a}{a-1}\)
x(x-3)-x2+9=0
x(x-3)-(x2-9)=0
x(x-3)-(x-3)(x+3)=0
(x-3)(x-x-3)=0
-3(x-3)=0
=> x=3
a) \(A=\frac{1-6x+9x^2}{x\left(3x-1\right)}=\frac{1^2-2.3x+\left(3x\right)^2}{x\left(3x-1\right)}=\frac{\left(1-3x\right)^2}{-x\left(1-3x\right)}=\frac{1-3x}{-x}=\frac{3x-1}{x}\)
b)
x2-y2+3x+3y=(x+y)(x-y)+3(x+y)=(x+y)(x-y-3)
\(x^2+2x+1-y^2=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
(x-2)2=x2-4x+4. Mình không hiểu cái câu này có ý nghĩa gì, 1 HĐT thôi mà
\(\frac{5x+3}{x+2}+\frac{7}{x+2}=\frac{5x+3+7}{x+2}=\frac{5x+10}{x+2}=\frac{5\left(x+2\right)}{x+2}=5\)
Đề này thì 10Đ khỏe
\(\frac{2a}{a-1}+\frac{a}{1-a}=\frac{2a}{a-1}+\frac{-a}{a-1}=\frac{2a-a}{a-1}=\frac{a}{a-1}\)
\(x\left(x-3\right)-x^2+9=0\Leftrightarrow x\left(x-3\right)-\left(x^2-9\right)=0\Leftrightarrow x\left(x-3\right)-\left(x+3\right)\left(x-3\right)=0\Leftrightarrow\left(x-x+3\right)\left(x-3\right)=0\Leftrightarrow3\left(x-3\right)=0=>x=3\)
