Question

Answer 1

toán mà để nhiều thế ko ai giải hết đâu tách ra

Answer 2

Bài 1 :

\(a.3x^3y-12x^2y^2+12xy^3\)

\(=3x^3y-6x^2y^2-6x^2y^2+12xy^3=3x^2y\left(x-2y\right)-6xy^2\left(x-2y\right)=3xy\left(x-2y\right)^2\)\(C2:3xy\left(x^2-4x^2y^2+4y^2\right)=3xy\left(x-2y\right)^2\)

Answer 3

\(b.x^3-x^2-4x+4=x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\)

\(c.x^2-9z^2-2xy+y^2=\left(x-y\right)^2-9z^2=\left(x-y-3z\right)\left(x-y+3z\right)\)

 

Answer 4

\(d.4x^2-9y^2+4x-6y=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x+3y+2\right)\left(2x-3y\right)\)\(e.x^2-2x+2y-y^2=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)

Answer 5

\(f.8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(2x+7\right)\left(4x+1\right)\)

Answer 6

\(Bài2:a.\left(-2x^2-4x+1\right)\left(2x-3x^2\right)=-4x^3-8x^2+12x^3+2x-3x^2+6x^{\text{4}}\)

\(=6x^4+8x^3-11x^2+2x\)

\(b.4x\left(x+1\right)+\left(3-2x\right)\left(3+2x\right)=4x^2+4x+9-4x^2=4x+9\)

 

Answer 7

\(c.\left(-3x^3+5x^2-9x+15\right):\left(5-3x\right)=\left[x^2\left(-3x+5\right)+3\left(-3x+5\right)\right]:\left(-3x+5\right)=\left(x^2+3\right)\left(-3x+5\right):\left(-3x+5\right)=x^2+3\)

Answer 8

\(d.\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)

\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

\(e.\dfrac{1}{x+1}-\dfrac{1}{x-1}+\dfrac{2x}{x^2-1}\)

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

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