6x2y3 + 3x2 - 10y3 = -2
ae giup mik nhé ^^
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a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)
b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)
c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)
a: \(3x^2-6xy+3y^2-12x^2\)
\(=3\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\left[\left(x-y\right)^2-4x^2\right]\)
\(=3\left(x-y-2x\right)\left(x-y+2x\right)\)
\(=3\left(-x-y\right)\left(3x-y\right)\)
b: \(3x^2y^2-6x^2y^3+12x^2y^2\)
\(=3x^2y^2\left(1-2y+4\right)\)
\(=3x^2y^2\left(-2y+5\right)\)
c: Ta có: \(3x^2-3y^2+12x-12y\)
\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
\(1,=3x^2-6x+x-2=3x^2-5x-2\\ 2,??\\ 3,=3x^3y^2:3xy+6x^2y^3:3xy-12xy^4:3xy=x^2y+2xy^2-4y^3\\ 4,=3x^3y^2:4xy+6x^2y^3:4xy-12xy^4:4xy\\ =\dfrac{3}{4}x^2y+\dfrac{3}{2}xy^2-3x^3\\ 5,\left(2x^3-5x^2+7x-6\right):\left(2x-3\right)=x^2-x+2\\ 6,\left(x^4-x^3+3x^2+x+2\right):\left(x^2-1\right)=x^2-x+4\left(dư6\right)\)
1: =3x^2+x-6x-2=3x^2-5x-2
3: =x^2y+2xy^2-4y^3
4: =3/4x^2y+3/2xy^2-3y^3
5: \(=\dfrac{2x^3-3x^2-2x^2+3x+4x-6}{2x-3}=x^2-x+2\)
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,
\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)
\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)
\(a,=18x^4-24x^3+30x\\ b,=3x^2y+6xy^2+x^2-6xy^2-12y^3-2xy=3x^2y+x^2-12y^3-2xy\\ c,=-3x^2+4xy-2x\\ d,=\left(x-y\right)^2\left[4\left(x-y\right)^3+2\left(x-y\right)-3\right]:\left(x-y\right)^2\\ =4\left(x-y\right)^3+2\left(x-y\right)-3\)
a: \(=18x^4-24x^3+30x^2\)
b: \(=3x^2y+6xy^2+x^2-6xy^2-12y^3-2xy\)
\(=x^2-12y^3+3x^2y-2xy\)
Câu 3:
a: 2x-8=4
nên 2x=12
hay x=6
b: 7x-3x=2x+7
\(\Leftrightarrow4x-2x=7\)
hay \(x=\dfrac{7}{2}\)
Câu 1:
a: \(5x\left(3x-4\right)=15x^2-20x\)
b: \(\left(x+5\right)\left(x-5\right)=x^2-25\)
\(\dfrac{-18x^4y^3-24x^3y^4+12x^3y^3}{-6x^2y^3}\)
\(=3x^2+4xy-2x\)