a. bc(b+c)+ca(c-a)-ab(a+b)
b. 2a^2b+4ab^2-a^2c+a^2-4b^2c+2bc^2-4abc
c. y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2
d. x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz
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2 tháng 9 2018
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
AH
Akai Haruma
Giáo viên
20 tháng 10 2020
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)
$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$
$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$
17 tháng 8 2020
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)
$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$
$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$
26 tháng 8 2018
a) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=\left(x+y+z\right)\left(xy+xz\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2+xyz\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z+x\right)\)
\(=\left(x+y+z\right)\left(xy+xz+yz\right)\)
P/s: Sai sót xin bỏ qua.
AH
Akai Haruma
Giáo viên
29 tháng 3 2020
Lời giải:
a)
$(a-x)y^3-(a-y)x^3+(x-y)a^3=(a-x)y^3-[(a-x)+(x-y)]x^3+(x-y)a^3$
$=(a-x)(y^3-x^3)+(x-y)(a^3-x^3)$
$=(a-x)(y-x)(y^2+xy+x^2)-(y-x)(a-x)(a^2+ax+x^2)$
$=(a-x)(y-x)(y^2+xy+x^2-a^2-ax-x^2)$
$=(a-x)(y-x)(y^2+xy-ax-a^2)=(a-x)(y-x)(y-a)(y+a+x)$
b)
$bc(b+c)+ca(c+a)+ba(a+b)+2abc$
$=bc(b+c+a)+ca(c+a+b)+ba(a+b)$
$=(bc+ac)(b+c+a)+ba(a+b)=c(b+a)(b+c+a)+ba(a+b)=(a+b)[c(a+b+c)+ab]$
$=(a+b)[c(a+c)+b(a+c)]=(a+b)(b+c)(c+a)$
c)
$x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz$
$=xy(x+y)+xz(x+z)+yz(y+z)+2xyz$
$=(x+y)(y+z)(x+z)$ (như phần b)
29 tháng 3 2020
c/\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
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