Violympic toán 8

DD

Phân tích các đa thức sau thành nhân tử:

a) \(yz.\left(y+z\right)+xz.\left(z-x\right)-xy.\left(x+y\right)\)

b) \(2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc\)

c) \(y.\left(x-2z\right)^2+8xyz+x.\left(y-2z\right)^2-2z.\left(x+y\right)^2\)

AH
20 tháng 10 2020 lúc 19:26

Lời giải:

a)

$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$

$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$

$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$

$=(y+z)(yz+xz-xy-x^2)$

$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$

b)

$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$

$=(2a^2b+4ab^2)-(a^2c+2abc)+(ac^2+2bc^2)-(4b^2c+2abc)$

$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$

$=(a+2b)(2ab-ac+c^2-2bc)$

$=(a+2b)[2b(a-c)-c(a-c)]$

$=(a+2b)(2b-c)(a-c)$

c)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$

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CD
17 tháng 8 2020 lúc 10:45

Lời giải:

a)

$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$

$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$

$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$

$=(y+z)(yz+xz-xy-x^2)$

$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$

b)

$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$

$=(2a^2b+4ab^2)-(a^2c+2abc)+(ac^2+2bc^2)-(4b^2c+2abc)$

$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$

$=(a+2b)(2ab-ac+c^2-2bc)$

$=(a+2b)[2b(a-c)-c(a-c)]$

$=(a+2b)(2b-c)(a-c)$

c)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$

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