Question

1)phân tích đa thức thành nhân tử

x²+xy+3x+3y

X²-y²+4x+4

X³+x-y-y³

2) 5/x+5-6/5-x+x²+25/x²-25

3x/x²-4+2/x-2+2/x+2

Answer 1

1)Phân tích đa thức thành nhân tử

\(a,x^2+xy+3x+3y\)

\(=x\left(x+y\right)+3\left(x+y\right)\)

\(=\left(x+y\right)\left(x+3\right)\)

\(b,x^2-y^2+4x+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2+y\right)\left(x+2-y\right)\)

\(c,x^3+x-y-y^3\)

\(=\left(x^3-y^3\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2+1\right)\)

2) \(\dfrac{5}{x+5}-\dfrac{6}{5-x}+\dfrac{x^2+25}{x^2-25}\)

\(=\dfrac{5}{x+5}+\dfrac{6}{x-5}+\dfrac{x^2+25}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{5\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{6\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{x^2+25}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{5x-25+6x+30+x^2+25}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+11x+30}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+5x+6x+30}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x+6\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x+6}{x-5}\)

\(3,\dfrac{x}{x^2-4}+\dfrac{2}{x-2}+\dfrac{2}{x+2}\)

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

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

\(=\dfrac{5x}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{x^2-4}\)

Answer 2

Answer 3

Answer 4

Answer 5

\(\left(x^2+xy\right)+\left(3x+3y\right)\\ =\left(x+y\right)\left(x+3\right)\\ x^2-y^2+4x+4\\ =\left(x^2+4x+4\right)-y^2\\ =\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)

\(x^3+x-y-y^3\\ =\left(x^3-y^3\right)+\left(x-y\right)\\ =\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\\ =\left(x-y\right)\left(x^2+xy+y^2+1\right)\)