Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

H24

giúp mình những bài toán này nhé

\(a.\left(x^2-y^2\right)-\left(5x+5y\right)\)

\(b.5x^3-5x^{2y}-10x^2+10xy\)

\(c.2x^2-5x\)

\(d.x^3-3x^2+1-3x\)

\(e.3x^2-6xy+3y^2-12z^2\)

\(F.3x^2-7x-10\)

\(g.x^4+1-2x^2\)

\(h.3x^2-3y^2-12x+12y\)

\(j.x^2-3x+2\)

NT
3 tháng 7 2017 lúc 10:59

a, \(\left(x^2-y^2\right)-\left(5x+5y\right)\)

\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

b, \(5x^3-5x^2y-10x^2+10xy\)

\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)

\(=\left(5x-10x\right)\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)

c, \(2x^2-5x=x\left(2x-5\right)\)

f, \(3x^2-7x-10=3x^2+3x^2-10x-10\)

\(=3x^2\left(x+1\right)-10\left(x+1\right)=\left(3x^2-10\right)\left(x+1\right)\)

d, \(x^3-3x^2+1-3x=x^3-3x^2-3x+1\)

\(=x^3+x^2-4x^2-4x+x+1\)

\(=x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)\)

\(=\left(x^2-4x+1\right)\left(x+1\right)\)

e, \(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-4z^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)

g, \(x^4+1-2x^2=\left(x^2-1\right)^2\)

h, \(3x^2-3y^2-12x+12y=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+3y-12\right)\)

\(=3\left(x-y\right)\left(x+y-4\right)\)

j, \(x^2-3x+2=x^2-2x-x+2=x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

Bình luận (2)
NS
3 tháng 7 2017 lúc 15:48

a. \(\left(x^2-y^2\right)-5\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-5\right)\)

b. \(5x^3-5x^2y-10x^2+10xy\)

\(=5\left[\left(x^3-x^2y\right)-\left(2x^2-2xy\right)\right]\)

\(=5\left[x^2\left(x-y\right)-2x\left(x-y\right)\right]\)

\(=5x\left(x-y\right)\left(x-2\right)\)

c. \(2x^2-5x=x\left(2x-5\right)\)

d. \(x^3-3x^2+1-3x\)

\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)

\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)

\(=\left(x+1\right)\left[x^2-x+1-3x\right]\)

\(=\left(x+1\right)\left[x^2-4x+1\right]\)

\(=\left(x+1\right)\left[x^2-2.x.2+2^2-2^2+1\right]\)

\(=\left(x+1\right)\left[\left(x-2\right)^2-3\right]\)

\(=\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)

e. \(3x^2-6xy+3y^2-12z^2\)

\(=3\left[x^2-2xy+y^2-4z^2\right]\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y+2z\right)\left(x-y-2z\right)\)

f. \(3x^2-7x-10\)

\(=3x^2-7x-7-3\)

\(=\left(3x^2-3\right)-\left(7x+7\right)\)

\(=3\left(x^2-1\right)-7\left(x+1\right)\)

\(=3\left(x+1\right)\left(x-1\right)-7\left(x+1\right)\)

\(=\left(x+1\right)\left[3\left(x-1\right)-7\right]\)

\(=\left(x+1\right)\left(3x-8\right)\)

g. \(x^4+1-2x^2=\left(x^2\right)^2-2.x^2+1=\left(x^2-1\right)^2\)

\(=\left(x+1\right)^2\left(x-1\right)^2\)

h. \(3x^2-3y^2-12x+12y\)

\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)

\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)

\(=\left(x-y\right)\left[3\left(x+y\right)-12\right]\)

\(=\left(x-y\right).3.\left(x+y-4\right)\)

j. \(x^2-3x+2=x^2-x-2x+2\)

\(=x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(x-2\right)\)

P/s: ( Có j sai ns nha nhiều số quá tui rối đầu )

Bình luận (2)
H24
4 tháng 7 2017 lúc 9:36

mình cảm ơn 2 bạn rất nhìu yeu

Bình luận (0)

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