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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!
a, 5x2 - 45x = 5x(x - 9)
b, 3x3y - 6x2y - 3xy3 - 6axy2 - 3a2xy + 3xy
= 3xy(x2 - 2x - y2 - 2ay - a2 + 1)
= 3xy[ (x2 - 2x + 1) - (a2 + 2ay + y2) ]
= 3xy[ (x - 1)2 - (a + y)2 ]
= 3xy(x - 1 + a + y)(x - 1 - a - y)
f, 3xy2 - 12xy + 12x
= 3x(y2 - 4y + 4)
= 3x(y - 2)2
g, 2x2 - 8x + 8
= 2(x2 - 4x + 4)
= 2(x - 2)2
h, 5x3 + 10x2y + 5xy2
= 5x( x2 + 2xy + y2 )
= 5x(x + y)2
k, x2 + 4x - 2xy - 4y + y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
i, x3 + ax2 - 4a - 4x
= (x3 - 4x) + (ax2 - 4a)
= x(x2 - 4) + a(x2 - 4)
= (x + a)(x2 - 4)
= (x + a)(x + 2)(x - 2)
Chúc bạn học tốt !
a) = -18x^4*y^4 + 3x^3*y^3 - 6/5x^2*y^4
b) = x^4 + 6x^3 + 9x^2 + 9x^2 - 6x^3 -18x^2 = x^4
bn dùng phương pháp nhân đơn thức vs đa thức đó dễ mà!!!!
chúc bạn học tốt!! ^^
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a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
Bài 1:
a) Sửa đề \(x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3y\right)\)
b) \(x^2+2019x-xy-2019y\)
\(=x\left(x+2019\right)-y\left(x+2019\right)\)
\(=\left(x+2019\right)\left(x-y\right)\)
c) \(x^2-9y^2-4x+4\)
\(=\left(x^2-4x+4\right)-9y^2\)
\(=\left(x-2\right)^2-\left(3y\right)^2\)
\(=\left(x-2-3y\right)\left(x-2+3y\right)\)
d) \(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
Bài 2:
a) \(\left(6x^3y^3-27xy^2\right):\left(3x^2y\right)-2xy^2\)
\(=6x^3y^3:3x^2y-27xy^2:3x^2y-2xy^2\)
\(=2xy^2-\dfrac{9y}{x}-2xy^2\)
\(=-\dfrac{9y}{x}\)
b) \(\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}+\dfrac{3x+2}{4-x^2}\)
\(=\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}-\dfrac{3x+2}{x^2-4}\)
\(=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(1-2x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)+\left(1-2x\right)\left(x-2\right)-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x+4+x-2-2x^2+4x-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x}{x+2}\)
Bài 3:
a) \(3x\left(2x-3\right)-x\left(6x+4\right)=7-12x\)
\(\Rightarrow6x^2-9x-6x^2-4x=7-12x\)
\(\Rightarrow-13x=7-12x\)
\(\Rightarrow-13x+12x-7=0\)
\(\Rightarrow-x-7=0\)
\(\Rightarrow-x=7\)
\(\Rightarrow x=-7\)
b) \(3\left(x-5\right)-2x^2+10x=0\)
\(\Rightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
a) \(=6x^3+8x^2+2x-6x^3=8x^2+2x\)
b) \(=\left[3xy\left(xy+2xy^2-4\right)\right]:3xy=xy+2xy^2-4\)
c) \(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x+2}-\dfrac{5}{x-2}=\dfrac{10x+3\left(x-2\right)-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)
a, \(=6x^3+12x^2+2x-6x^3\\=12x^2+2x\)
b,
\(=xy+2xy^2-4\)
c,
\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)
\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)