Thực hiện phép tính:
(x + 3) (x - 3) - x^2
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Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)
\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)
\(=\dfrac{x^2+2x-3x+6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2\left(x-2\right)\left(x+2\right)}{x-3}\)
\(=\dfrac{2\left(x^2-x+6\right)}{x-3}\)
\(\left(x^3-x^2-5x-3\right):\left(x-3\right)\\ =\left[\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(x-3\right)\right]:\left(x-3\right)\\ =\left[x^2\left(x-3\right)+2x\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)\\ =\left[\left(x-3\right)\left(x^2+2x+1\right)\right]:\left(x-3\right)\\ =x^2+2x+1\)
giúp tui với mn ơi
\(=x^2-9-x^2=-9\)