Toán Lớp 8: Phân tích các đa thức sau thành nhân tử: a, 3xy+x+15y+5 b, 5xy-5xz+25y-25z c, 13x+13y-x^2-xy d, x^2-xy-8x+8y e, 5xy-10xz+22-y g, x^3-2x

Question

Toán Lớp 8:  Phân tích các đa thức sau thành nhân tử: a, 3xy+x+15y+5 b, 5xy-5xz+25y-25z c, 13x+13y-x^2-xy d, x^2-xy-8x+8y e, 5xy-10xz+22-y g, x^3-2x^2-9x+18 h, 6x^2+12xa+6a^2-24b^2 i, 9x^2-y^2+2yz-z^2

truongthanhhung · Answer

Giải đáp:     a. 3xy+x+15y+5=x(3y+1)+5(3y+1)=(3y+1)(x+5)   b. 5xy-5xz+25y-25z=5(xy-xz+5y-5z)=5(x(y-z)+5(y-z))=5(y-z)(x+5)   c. 13x+13y-x^2-xy=13(x+y)-x(x+y)=(x+y)(13-x)   d. x^2-xy-8x+8y=x(x-y)-8(x-y)=(x-y)(x-8)   e. 5xy-10xz+2z-y=5xy-10xz-y+2z=5x(y-2z)-(y-2z)=(y-2z)(5x-1)   g. x^3-2x^2-8x+18=x^2(x-2)-8(x-2)=(x-2)(x^2-8)   h. 6x^2+12xa+6a^2-24b^2=6(x^2+2xa+a^2-4b^2)=6((x+a)^2-4b^2)=6(a+x-2b)(x+a+2b)   i. 9x^2-y^2+2yz-z^2=9x^2-(y-z)^2=(3x-y+z)(3x+y-z)

kimoanh34 · Answer

Giải đáp:   a.$(x+5)(3y+1)$   b.$5(y-z)(x+5)$   c.$(13-x)(x+y)$   d.$(x-8)(x-y)$   e.$(5x-1)(y-2z)$   g.$(x-3)(x+3)(x-2)$   h.$6(x+a-2b)(x+a+2b)$   i.$(3x-y+z)(3x+y-z)$   Lời giải và giải thích chi tiết:   a.Ta c&oacute;:   $3xy+x+15y+5$   $=(3xy+x)+(15y+5)$   $=x(3y+1)+5(3y+1)$   $=(x+5)(3y+1)$   b.Ta c&oacute;:   $5xy-5xz+25y-25z$   $=(5xy-5xz)+(25y-25z)$   $=5x(y-z)+25(y-z)$   $=5(y-z)(x+5)$   c.Ta c&oacute;:   $13x+13y-x^2-xy$   $=(13x+13y)-(x^2+xy)$   $=13(x+y)-x(x+y)$   $=(13-x)(x+y)$   d.Ta c&oacute;:   $x^2-xy-8x+8y$   $=(x^2-xy)-(8x-8y)$   $=x(x-y)-8(x-y)$   $=(x-8)(x-y)$   e.Ta c&oacute;:   $5xy-10xz+2z-y$   $=(5xy-10xz)+(2z-y)$   $=5x(y-2z)-(y-2z)$   $=(5x-1)(y-2z)$   g.Ta c&oacute;:   $x^3-2x^2-9x+18$   $=(x^3-2x^2)-(9x-18)$   $=x^2(x-2)-9(x-2)$   $=(x^2-9)(x-2)$   $=(x-3)(x+3)(x-2)$   h.Ta c&oacute;:   $6x^2+12xa+6a^2-24b^2$   $=(6x^2+12xa+6a^2)-24b^2$   $=6(x^2+2xa+a^2)-24b^2$   $=6(x+a)^2-24b^2$   $=6((x+a)^2-4b^2)$   $=6((x+a)^2-(2b)^2)$   $=6(x+a-2b)(x+a+2b)$   i.Ta c&oacute;:   $9x^2-y^2+2yz-z^2$   $=9x^2-(y^2-2yz+z^2)$   $=(3x)^2-(y-z)^2$   $=(3x-(y-z))(3x+(y-z))$   $=(3x-y+z)(3x+y-z)$