Rút gọn (-3.x3y)2(-2xy)3
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\(\dfrac{8x^3+y^3}{y^3+2xy^2+y^2-4x^2}\\ =\dfrac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{\left(y^3+2xy^2\right)+\left(y^2-4x^2\right)}\\ =\dfrac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{y^2\left(y+2x\right)+\left(y-2x\right)\left(y+2x\right)}\\ =\dfrac{\left(y+2x\right)\left(4x^2-2xy+y^2\right)}{\left(y+2x\right)\left(y^2+y-2x\right)}\\ =\dfrac{4x^2-2xy+y^2}{y^2+y-2x}\)
\(\dfrac{8x^3+y^3}{y^3+2xy^2+y^2-4x^2}\)
\(=\dfrac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{y^2\left(y+2x\right)+\left(y+2x\right)\left(y-2x\right)}\)
\(=\dfrac{4x^2-2xy+y^2}{y^2+y-2x}\)
a) Ta có: \(\left(y+3\right)\left(y^2-3y+9\right)-\left(60-y^3\right)\)
\(=y^3+27-60+y^3\)
\(=2y^3-33\)
b) Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(\dfrac{2xy-x^2}{3x^3-6x^2y}\\ =\dfrac{x\left(2y-x\right)}{3x^2\left(x-2y\right)}\\ =\dfrac{x\left(2y-x\right)}{-3x^2\left(2y-x\right)}\\ =\dfrac{1}{-3x}\)
\(\dfrac{2xy-x^2}{3x^3-6x^2y}=\dfrac{-\left(x^2-2xy\right)}{3x^3-6x^2y}\)
\(=\dfrac{-x\left(x-2y\right)}{3x^2\left(x-2y\right)}=\dfrac{-1}{3x}\)
a) (x+3)(x^2-3x+9)-(54+x^3)
= x^3- 3x^2+9x+3x^2-9x+27-54-x63
= -27
b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)
= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]
= [(2x)3^3+ y^3] – [(2x)^3 – y^3]
= (2x)^3 + y^3 – (2x)^3 + y^3
= 2y^3
a)(x+3)(X^2-3x+9)-(54+x^3)
= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)
= 27- 54
= -27
b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
= \((2x)^3\) + \(y^3\) - [\((2x)^3\) - \(y^3\) ]
= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)
= \(2y^3\)
Bài làm
\(\frac{4x^3y^2-6x^2y^3}{2xy+2xy\left(y-x\right)}=\frac{2x^2y^2\left(2x-3y\right)}{2xy\left(1+y-x\right)}=\frac{xy\left(2x-3y\right)}{1+y-x}\)
Học tốt
\(\frac{4x^3y^2-6x^2y^3}{2xy+2xy\left(y-x\right)}=\frac{2x^2y^2\left(2x-3y\right)}{2xy\left(1+y-x\right)}=\frac{xy\left(2x-3y\right)}{y-x+1}\)