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a, (-2x5+3x2-4x3):2x2=-x3+3/2-2x=-x3-2x+3/2
b,(x3-2x2y+3xy2):(-1/2.x)=-2x2+4xy-6y2
c,(3x2y2+6x2y3-12xy):3xy=xy+2xy2-4
a) -2x5 + 3x2 - 4x3 = x2. (-2x3 + 3 - 4x) => (-2x5 + 3x2 - 4x3 ) : 2x2 = [ x2. (-2x3 + 3 - 4x) ] : (2x2) = (-2x3 + 3 - 4x) /2 = -x3 - 2x + (3/2)
b) x3 - 2x2y + 3xy2 = x. (x2 - 2xy + 3y2)
=> x3 - 2x2y + 3xy2 : (-1/2. x) = (x2 - 2xy + 3y2) : (-1/2) = (x2 - 2xy + 3y2) . (-2) =-2 x2 + 4xy -6y2
c) 3x2y2 + 6x2y3 - 12xy = 3xy. (xy + 2xy2 - 4)
=> 3x2y2 + 6x2y3 - 12xy : 3xy = xy + 2xy2 - 4
a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)\)
\(=x^2+y^2\)
b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)
\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)
\(=7x-9y\)
c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)
\(=\left(x+y\right)^3:\left(x+y\right)\)
\(=\left(x+y\right)^2=x^2+2xy+y^2\)
d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3:\left(x-y\right)^2\)
\(=\left(x-y\right)\)
e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)
Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)
\(=\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{2x+1}\)
\(=4x^2-2x+1\)
f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\frac{\left(2x-1\right)\left(4x^2+2x+1\right)}{4x^2+2x+1}\)
\(=2x-1\)
a, (x4 + 2x2y2 + y4) : (x2 + y2)
= (x2 + y2)2 : (x2 + y2)
= x2 + y2
b, (49x2 - 81y2) : (7x + 9y)
= (7x - 9y)(7x + 9y) : (7x + 9y)
= 7x - 9y
c, (x3 + 3x2y + 3xy2 + y3) : (x + y)
= (x + y)3 : (x + y)
= (x + y)2
d, (x3 - 3x2y + 3xy2 - y3) : (x2 - 2xy + y2)
= (x - y)3 : (x - y)2
= x - y
Phần e thiếu thì phải
f, (8x3 - 1) : (4x2 + 2x + 1)
= (2x - 1)(4x2 + 2x + 1) : (4x2 + 2x + 1)
= 2x - 1
Chúc bn học tốt!
a: \(=\dfrac{x^2-x+1-4x}{xy}=\dfrac{x^2-5x+1}{xy}\)
b: \(=\dfrac{5xy^2-x^2y+4xy^2+xy^2}{3xy}\)
\(=\dfrac{10xy^2-x^2y}{3xy}=\dfrac{xy\left(10y-x\right)}{3xy}=\dfrac{10y-x}{3}\)
d: \(\dfrac{2x+4}{10}-\dfrac{2-x}{15}\)
\(=\dfrac{x+2}{5}+\dfrac{x-2}{15}\)
\(=\dfrac{3x+6+x-2}{15}=\dfrac{4x+4}{15}\)
e: \(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)
a)\(\frac{2x^2y-3xy}{4x^2+x}=\frac{xy\left(2x-3\right)}{x\left(4x+1\right)}\)\(=\frac{y\left(2x-3\right)}{4x+1}\)
b)\(\frac{3x^2+x-2}{x+1}=\frac{\left(x+1\right)\left(3x-2\right)}{x+1}=3x-2\)