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Bài 2:
\(=\dfrac{x^2\left(x^2+4\right)-2x\left(x^2+4\right)}{x^2+4}=x^2-2x\)
Bài 1:
a: \(=\left(\dfrac{2}{3}:\dfrac{-1}{9}\right)\cdot x^4y^2z^6=-6x^4y^2z^6\)
b: \(=-12x^8-21x^5\)
c: =x^3+8
d: \(=125x^3-75x^2+15x-1\)
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a) \(\dfrac{6x^2y^3-2x^2y+6xy}{6xy}\)
\(=\dfrac{6x^2y^3}{6xy}-\dfrac{2x^2y}{6xy}+\dfrac{6xy}{6xy}\)
\(=xy^2-\dfrac{x}{3}+1\)
b) \(\dfrac{4\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{2\left(x+y\right).2\left(x+y\right)^2}{2\left(x+y\right)}\)
\(=2\left(x+y\right)^2\)
c) \(\dfrac{8x^3+27y^3}{2x+3y}\)
\(=\dfrac{\left(2x\right)^3+\left(3y\right)^3}{2x+3y}\)
\(=\dfrac{\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]}{2x+3y}\)
\(=4x^2-6xy+9y^2\)
d) \(\dfrac{48x^4y^3-12x^2y^5+6x^2y^2}{3x^2y^2}\)
\(=\dfrac{48x^4y^3}{3x^2y^2}-\dfrac{12x^2y^5}{3x^2y^2}+\dfrac{6x^2y^2}{3x^2y^2}\)
\(=16x^2y-4y^3+2\)
1) \(-4x^5\left(x^3-4x^2+7x-3\right)\)
\(=-4x^8+16x^7-28x^6+12x^5\)
2) \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=-6x^7+15x^6-2x^5+x^4\)
3) \(-5x^2y^4\left(3x^2y^3-2x^3y^2-xy\right)\)
\(=-15x^4y^7+10x^5y^6+5x^3y^5\)
4) \(4x^3y^2\left(-2x^2y+4x^4-3y^2\right)\)
\(=-8x^5y^3+16x^7y^2-12x^3y^4\)
a) \(\left(8x^5-4x^3\right):\left(-2x^3\right)\)
\(=\left[8x^5:\left(-2x^3\right)\right]+\left[\left(-4x^3\right):\left(-2x^3\right)\right]\)
\(=-4x^2+2\)
b) \(\left(-2x^2y^3-3x^2y^2+x^2y\right):\left(\dfrac{-1}{3}x^2y\right)\)
\(=\left[\left(-2x^2y^3\right):\left(\dfrac{-1}{3}x^2y\right)\right]+\left[\left(-3x^2y^2\right):\left(\dfrac{-1}{3}x^2y\right)\right]+\left[x^2y:\left(\dfrac{-1}{3}x^2y\right)\right]\)
\(=6y^2+9y+\left(-3\right)\)
\(=6y^2+9y-3\)