Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải
1) 3xy2 : 5x = \(\frac{3}{5}\)y2
2) 15x4yz3 : 4xyz = \(\frac{15}{4}\)x3z2
3) (4x2y2 - 12xy3 - 7x) : 3x = \(\frac{4}{3}\)xy2 - 4y3 - \(\frac{7}{3}\)
4) (14x4y2 - 12xy3 - x) : 4x = \(\frac{7}{2}\)x3y2 - 3y3 - \(\frac{1}{4}\)
5) (6x2 + 13x - 5) : (2x + 5) = (3x - 1)(2x + 5) : (2x + 5) = 3x - 1
6) (2x4 + x3 - 5x2 - 3x - 3) : (x2 - 3)
= 2x4 + x2 - 6x2 + x3 - 3 - 3x : x2 - 3
= x2(2x2 + x + 1) - 3(2x2 + x + 1) : x2 - 3
= (2x2 + x + 1)(x2 - 3) : x2 - 3
= 2x2 + x + 1
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
1: \(=\dfrac{x^2\cdot4xy^2}{x^2}=4xy^2\)
2: \(=\dfrac{3x\left(x-2\right)}{-\left(x-2\right)}=-3x\)
3: \(=\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x^2+2x+4}=x-2\)
6: \(\dfrac{5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2}{\left(x-y\right)^2}=5\left(x-y\right)^2-3\left(x-y\right)+4\)
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!
