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a: \(=x^2+2x-8-x^2-2x-1=-9\)
b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
Bài 3:
\(\Leftrightarrow x^3+64-x^3+25x=264\)
hay x=8
\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)
\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)
b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x^2-3x}\)
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)
\(=x^3-8-\left(x^3-3x^2+3x-1\right)+7\)
\(=x^3-8+7-x^3+3x^2-3x+1\)
\(=\left(x^3-x^3\right)+\left(7+1-8\right)+3x^2-3x\)
\(=3x^2-3x=3x\left(x-1\right)\)
\(x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x\left(2+x\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=x\left(4-x^2\right)+\left(x+3\right)\left(x^2-3x+9\right)\)
\(=4x-x^3+\left(x^3+9\right)\)
\(=4x-\left(x^3-x^3\right)+9\)
\(=4x+9\)
\(\dfrac{x+3}{x^2-3x}+\dfrac{3}{x^2+3x}+\dfrac{2x-18}{x^2-9}\\ =\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
\(=\dfrac{x+3}{x\left(x-3\right)}+\dfrac{3}{x\left(x+3\right)}+\dfrac{2x-18}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+6x+9+3x-9+4x-36}{x\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+13x-36}{x\left(x+3\right)\left(x-3\right)}\)