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mình k cho bạn rồi nha, tích lại cho mình, số điểm của mình là -159 điểm
a)x2-y2-5x+5y
=(x-y)(x+y)-5.(x-y)
=(x-y)(x+y-5)
b)5x3-5xy-10x2+10xy
=5x3+5xy-10x2
=5x.(x2+y-2x)
c)x3-3x2+1-3x
=x3+1-3x2-3x
=(x+1)(x2-x+1)-3x.(x+1)
=(x+1)(x2-x+1-3x)
=(x+1)(x2-4x+1)
d)3x2-6xy+3y2-12z2
=3.(x2-2xy+y2-4z2)
=3.[(x-y)2-4z2]
=3.(x-y-2z)(x-y+2z)
3(x2-2xy+y2):10(x-y)
3(x-y)2:10(x-y)
3(x-y):10
5(x2-xy+y2):(x+y)(x2-xy+y2)
5:x+y
ở nơi 5y thiếu ^2 nhé
bài 1
a) ta có: \(8x^3+12x^2y-2xy^2-3y^3\)
\(=\left(8x^3+12x^2y\right)-\left(2xy^2+3y^3\right)\)
\(=4x^2\left(2x+3y\right)-y^2\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(4x^2-y^2\right)\)
\(=\left(2x+3y\right)\left(2x-y\right)\left(2x+y\right)\)
a) \(x^2-y^2-5x-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b) \(5x^3-5x^2y-10x^2+10xy\)
\(=\left(5x^3-5x^2y\right)-\left(10x^2-10xy\right)\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x^2-10x\right)\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c) \(x^3-2x^2-x+2\)
\(=\left(x^3-2x^2\right)-\left(x-2\right)\)
\(=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(-y^2+2xy-x^2+3x-3y\)
\(=-\left(y^2-2xy+x^2\right)+\left(3x-3y\right)\)
\(=-\left(y-x\right)^2+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
g) \(4x^2-8x+3\)
\(=4x^2-6x-2x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-1\right)\)
h) \(2x^2-5x-7\)
\(=2x^2+2x-7x-7\)
\(=\left(2x^2+2x\right)-\left(7x+7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(2x-7\right)\)
k) \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
\(\left(2x^2-y\right)\left(4x^2-5xy^2+3y^2\right)\)
\(=\left(2x^2-y\right)4x^2-\left(2x^2-y\right)5xy^2+\left(2x^2-y\right)3y^2\)
\(=8x^4-4x^2y-10x^3y^2+5xy^3+6x^2y^2-3y^3\)
\(\text{Câu 1: }\left(2x^2-y\right)\left(4x^2-5xy^2+3y^2\right)\\ \\=2x^2\left(4x^2-5xy^2+3y^2\right)-y\left(4x^2-5xy^2+3y^2\right)\\ \\=\\8x^4-10x^3y^2+6x^2y^2-4x^2y+5xy^3+3y^3\)
Câu 2:
\(\text{ a) }48x^2y^2-3y^2+6xy-3x^2\\ \\ =3\left(16x^2y^2-y^2+2xy-x^2\right)\\ \\ =3\left[16x^2y^2-\left(x^2-2xy+y^2\right)\right]\\ =3\left[\left(4xy\right)^2-\left(x-y\right)^2\right]\\ \\ =3\left(4xy-x+y\right)\left(4xy+x-y\right)\)
\(\text{b) }2x^3y-4x^2y^2+2xy^3\\ \\=2xy\left(x^2-2xy+y^2\right)\\ \\=2xy\left(x-y\right)^2\)
\(\text{c) }4x^2-6x^3y-2x^2+8x\\ \\=2x^2-6x^3y+8x\\ \\ =2x\left(x-3x^2y+4\right)\)
\(\text{d) }6xy+5x-5y-3x^2-3y^2\\ \\ =\left(5x-5y\right)-\left(3x^2-6xy+3y^2\right)\\ \\ =5\left(x-y\right)-3\left(x^2-2xy+y^2\right)\\ \\ =5\left(x-y\right)-3\left(x-y\right)^2\\ \\ =\left(x-y\right)\left[5-3\left(x-y\right)\right]\\ =\left(x-y\right)\left(5-3x+3y\right)\)
giup minh voi, chieu minh thi roi @@
=(xy(8x+4+5y))/2xy -3x^2
=(8x+4+5y)/2 +3x^2
=(8x+4+5y)/2 + 6x^2 /2
=(8x+4+5y-6x^2)/2