x^2 - 3x - 4+ 6xy + 9y^2+ 9y
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\(x^4+6x^3+13x^2+12x+4\)
\(=x^4+x^3+5x^3+5x^2+8x^2+8x+4x+4\)
\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+8x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+5x^2+8x+4\right)\)
\(=\left(x+1\right)\left(x^3+x^2+4x^2+4x+4x+4\right)\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\right]\)
\(=\left(x+1\right)^2\left(x+2\right)^2\)
a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{x+2}\)
\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)
b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)
\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)
=0
x^2+6xy+9y^2-3x-9y+2
=( x^2+6xy+9y^2)-3(x+3y)+9/4 -1/4
=(x+3y)^2-3(x+3y)+(3/2)^2- 1/4
=(x+3y+3/2)^2-(1/2)^2
=(x+3y+3/2+1/2)(x+3y+3/2-1/2)=(x+3y+2)(x+3y+1)
\(=\dfrac{4}{\left(x-3y\right)\left(x+3y\right)}\cdot\pm\left(2\sqrt{2}\right)\cdot y\cdot\left(x-3y\right)\)
\(=\dfrac{\pm8\sqrt{2}\cdot y}{x+3y}\)