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a) \(x^3\left(3x^2-x-\dfrac{1}{2}\right)=3x^5-x^4-\dfrac{1}{2}x^3\)
b) \(\left(5xy-x^2+y\right)\dfrac{2}{5}xy^2=2x^2y^3-\dfrac{2}{5}x^3y^2+\dfrac{2}{5}xy^3\)
c) \(\left(4x^3-3xy^2+2xy\right)\left(-\dfrac{1}{3}x^2y\right)=\dfrac{-4}{3}x^5y+x^3y^3-\dfrac{2}{3}x^3y^2\)
a ) có \(x^2+y^2+4x-2xy+4y+2019=\left(x-y\right)^2+4\left(x-y\right)+2019=49+28+2019=2096\)
b) \(x^3-3xy\left(x-y\right)-y^3-x^2+2xy-y^2=\left(x-y\right)^3-\left(x-y\right)^2=343-49=294\)
c)\(x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)=x^3-y^3+x^2+y^2+xy-3x^2y+3xy^2-3xy=\left(x-y\right)^3+\left(x-y\right)^2=343+49=392\)
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) x2(5x3 – x - 1212) = x2. 5x3 + x2 . (-x) + x2 . (-1212)
= 5x5 – x3 – 1212x2
b) (3xy – x2 + y) 2323x2y = 2323x2y . 3xy + 2323x2y . (- x2) + 2323x2y . y
= 2x3y2 – 2323x4y + 2323x2y2
c) (4x3– 5xy + 2x)(- 1212xy) = - 1212xy . 4x3 + (- 1212xy) . (-5xy) + (- 1212xy) . 2x
= -2x4y + 5252x2y2 - x2y.
\(a,\left(4x^3-3xy^2+2xy\right).\left(-\dfrac{1}{3}x^2y\right)\)
\(=\dfrac{-x}{3y}+\dfrac{y}{x}-\dfrac{2}{3x}\)
\(b,\left(5xy-x^2+y\right)\left(\dfrac{2}{5}xy^2\right)\)
\(=\dfrac{2}{y}-\dfrac{2x}{5y^2}+\dfrac{2}{5xy}\)
c,=\(\dfrac{-4}{3}x^5y+x^3y^3-\dfrac{2}{3}x^3y^2\) b,=\(2x^2y^3-\dfrac{2}{5}x^3y^2+\dfrac{2}{5}xy^3\)
\(b,\left(5xy-x^2+y\right).\dfrac{2}{5xy^2}\)
\(=5xy.\dfrac{2}{5xy^2}-x^2.\dfrac{2}{5xy^2}+y.\dfrac{2}{5xy^2}\)
\(=\dfrac{2}{y}-\dfrac{2x}{5y^2}+\dfrac{2}{5xy}\)
\(c,\left(4x^3-3xy^2+2xy\right)\left(-\dfrac{1}{3x^2y}\right)\)
\(=4x^3.\left(\dfrac{-1}{3x^2y}\right)-3xy^2.\left(\dfrac{-1}{3x^2y}\right)+2xy\left(-\dfrac{1}{3x^2y}\right)\)\(=\dfrac{-x}{3y}+\dfrac{y}{x}-\dfrac{2}{3x}\)
Yêu cầu bài là gì vậy?