y+x2y+3x2+3
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\(3x^2\left(2x^2-y\right)-4x^2\left(x^2y-y^2\right)\)
\(=6x^4-3x^2y-4x^4y+4x^2y^2\)
a) M - \(^{\left(x^2y-1\right)}\)= -2\(x^3\)+\(x^2y\)+1
=> M= (-2\(x^3\)+\(x^2y\)+1) + \(^{\left(x^2y-1\right)}\)
=> M= -2\(x^3\)+\(x^2y\)+1+ \(^{x^2y-1}\)
=> M= -2\(x^3\)+(\(x^2y+x^2y\))+1-1
=> M= -2\(x^3\)+\(2x^2y\)
b) \(3x^2+3xy-3x^3-M=3x^2+2xy-4y^2\)
=> \(M=\left(3x^2+3xy-3x^3\right)-\left(3x^2+2xy-4y^2\right)\)
\(=>M=3x^2+3xy-3x^3-3x^2-2xy+4y^2\)
\(=>M=\left(3x^2-3x^2\right)+\left(3xy-2xy\right)-3x^3+4y^2\)
\(=>M=xy-3x^3+4y^2\)
Hơi muộn nhưng mong bạn tick cho mình
a, 2xy +2x2 - 4xy2 - 2 ; b, -3x2y2 -2x2y + y ; c, 3x3 - 2y - 3
a) (3x2 – 2x2y) : x2 – (2xy2 + x2y) : (1/3 xy)
= (3x3 : x2) + (-2x2y : x2) - [(2x2y : 1/3 xy) +( x2y : 1/3 xy)]
= 3x – 2y – (6y + 3x) = 3x – 2y – 6y – 3x = -8y
Ta có A + 2B = (x2y - xy2 + 3x2) + 2(x2y + xy2 - 2x2 - 1)
= x2y - xy2 + 3x2 + 2x2y + 2xy2 - 4x2 - 2
= 3x2y + xy2 - x2 - 2. Chọn C
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
\(a,=3x^3y^3-3x^2y^3+3x^2y^4+3xy^5\\ b,=\left(2x^3-6x^2+10x-3x^2+9x-15\right):\left(x^2-3x+5\right)\\ =\left[2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)\right]:\left(x^2-3x+5\right)\\ =2x-3\\ c,=\left[x^2\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)=x^2+1\)
\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)
Lời giải:
a.
\(\frac{x^2+xy+x+y}{x^2-xy+x-y}=\frac{x(x+y)+(x+y)}{x(x+1)-y(x+1)}=\frac{(x+y)(x+1)}{(x+1)(x-y)}=\frac{x+y}{x-y}\)
b.
\(\frac{x^2-6x+9}{3x^2-9x}=\frac{(x-3)^2}{3x(x-3)}=\frac{x-3}{3x}\)
c.
\(\frac{y^2-x^2}{x^2y-xy^2}=\frac{(y-x)(y+x)}{-xy(y-x)}=\frac{x+y}{-xy}\)
\(y+x^2y+3x^2+3=\left(y+x^2y\right)+\left(3x^2+3\right)=y\left(x^2+1\right)+3\left(x^2+1\right)=\left(x^2+1\right)\left(y+3\right)\)
\(y+x^2y+3x^2+3\)
\(=y\left(x^2+1\right)+3\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(y+3\right)\)