12x3y4+9x3y3-6xy2:3xy
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\(Sửa:\left(9x^2y-6xy^2+5xy\right):3xy=3x-2y+\dfrac{5}{3}\)
a)=\(3x^3-15x^2+21x\)
b)\(=-2x^4y-10x^2y+2xy\)
c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)
d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)
e)\(=x^2-4y^2\)
f)\(=-2x^2y^3+y-3\)
g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)
h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)
i)\(=x^2-x-3\)
j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)
\(a,3\left(x^2-7\right)-x\left(3x+5\right)=3x^2-21-3x^2-5x=-5x-21\\ b,\left(12x^2y^2-6xy\right):3xy+2y=3xy\left(4xy-2\right):3xy+2y=4xy-2+2y\)
\(c,\dfrac{4}{x+1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x-1\right)+8}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x-4+8}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\)
Ta có
( 9 x 4 y 3 – 18 x 5 y 4 – 81 x 6 y 5 ) : ( - 9 x 3 y 3 ) = [ ( 9 x 4 y 3 ) : ( - 9 x 3 y 3 ) ] – [ 18 x 5 y 4 : ( - 9 x 3 y 3 ) ] – [ 81 x 6 y 5 : ( - 9 x 3 y 3 ) ] = - x + 2 x 2 y + 9 x 3 y 2
Đa thức - x + 2 x 2 y + 9 x 3 y 2 có bậc 3 + 2 = 5
Đáp án cần chọn là: A
\(a,=12x^2+18x-12x^2+x+1=19x+1\\ b,=4x^3y^4-3xy^2-\dfrac{3}{2}x\)
\(a.x\left(2x+5\right)=2x^2+5x\)
\(b.3x^2\left(5x^3-2x+9\right)=15x^5-6x^3+27x^2\)
\(c.\left(x+2\right)\left(4x^3+x-7\right)=4x^4+x^2-7x+8x^3+2x-14=4x^4+8x^3+x^2-5x-14\)
\(d.\left(16x^2+12x^3y^4-4x^2y\right):4x^2y=4x^3+3x^2y^3-1\)
`(12x^3y^4 + 9x^3y^3 - 6xy^2) : 3xy`
`= 4x^2y^3 + 3x^2y^2 - 2y`.