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\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
a/ x3 + x2 z + y2 z - xyz + y3
= (x + y)(x2 - xy + y2) + z(x2 - xy + y2)
= (x2 - xy + y2)(x + y + z)
* Đặt tên các biểu thức theo thứ tự là A,B,C,D,E.
Câu a)
Theo hằng đẳng thức đáng nhớ ta có:
\(a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)\)
\(=(a+b+c)^3-3[ab(a+b)+bc(b+c)+ca(c+a)+2abc]\)
\(=(a+b+c)^3-3[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)-abc]\)
\(=(a+b+c)^3-3[(a+b+c)(ab+bc+ac)]+3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)^3-3(ab+bc+ac)(a+b+c)\)
\(=(a+b+c)[(a+b+c)^2-3(ab+bc+ac)]\)
\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\) (*)
Do đó:
\(A=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)
Câu b)
\(x^3-y^3+z^3+3xyz=x^3+(-y)^3+z^3-3x(-y)z\)
Sử dụng kết quả (*) của câu a. Với \(a=x, b=-y, c=z\)
\(\Rightarrow x^3+(-y)^3+z^3-3x(-y)z=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)
Mặt khác xét mẫu số:
\((x+y)^2+(y+z)^2+(x-z)^2=x^2+2xy+y^2+y^2+2yz+z^2+x^2-2xz+z^2\)
\(=2(x^2+y^2+z^2+xy+yz-xz)\)
Do đó: \(B=\frac{(x-y+z)(x^2+y^2+z^2+xy+yz-xz)}{2(x^2+y^2+z^2+xy+yz-xz)}=\frac{x-y+z}{2}\)
Câu c) Sử dụng kết quả (*) của phần a:
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\)
Và mẫu số:
\((x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2-xy-yz-xz)\)
Do đó: \(C=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{2(x^2+y^2+z^2-xy-yz-xz)}=\frac{x+y+z}{2}\)
Câu d)
Xét tử số:
\(a^2(b-c)+b^2(c-a)+c^2(a-b)\)
\(=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)\)
\(=(b-c)(a^2-b^2)-(b^2-c^2)(a-b)\)
\(=(b-c)(a-b)(a+b)-(b-c)(b+c)(a-b)\)
\(=(a-b)(b-c)[a+b-(b+c)]=(a-b)(b-c)(a-c)\) (1)
Xét mẫu số:
\(a^4(b^2-c^2)+b^4(c^2-a^2)+c^4(a^2-b^2)\)
\(=a^4(b^2-c^2)-b^4[(b^2-c^2)+(a^2-b^2)]+c^4(a^2-b^2)\)
\(=(a^4-b^4)(b^2-c^2)-(b^4-c^4)(a^2-b^2)\)
\(=(a^2-b^2)(a^2+b^2)(b^2-c^2)-(b^2-c^2)(b^2+c^2)(a^2-b^2)\)
\(=(a^2-b^2)(b^2-c^2)[a^2+b^2-(b^2+c^2)]\)
\(=(a^2-b^2)(b^2-c^2)(a^2-c^2)\)
\(=(a-b)(b-c)(a-c)(a+b)(b+c)(c+a)\)(2)
Từ (1)(2) suy ra \(D=\frac{1}{(a+b)(b+c)(c+a)}\)
Câu e)
Theo phần d ta có:
\(TS=(a-b)(b-c)(a-c)\)
\(MS=ab^2-ac^2-b^3+bc^2\)
\(=b^2(a-b)-c^2(a-b)=(a-b)(b^2-c^2)=(a-b)(b-c)(b+c)\)
Do đó: \(E=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(b+c)}=\frac{a-c}{b+c}\)
2) Để sau đi (em chưa nghĩ ra)
3) \(A=\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)+\left(y+z\right)^2\left(y-z\right)+\left(z+x\right)^2\left(z-x\right)\)
Đặt x - y = a; y - z = b => z - x = -(a+b)
\(A=\left(x+y\right)^2a+\left(y+z\right)^2b-\left(z+x\right)^2a-\left(z+x\right)^2b\)
\(=a\left[\left(x+y\right)^2-\left(z+x\right)^2\right]+b\left[\left(y+z\right)^2-\left(z+x\right)^2\right]\)
\(=\left(x-y\right)\left(x+y-z-x\right)\left(x+y+z+x\right)+\left(y-z\right)\left(y+z-z-x\right)\left(y+z+z+x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(2x+y+z\right)-\left(y-z\right)\left(x-y\right)\left(2z+x+y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Em tính sai sót chỗ nào thì thông cảm cho em ạ :>
1)
=2(a4+b4+c4-4a2b2-4a2c2-4b2c2)
=2a4+2b4+2c4-4a2b2-4a2c2-4b2c2
=(a4-2a2b2+b4)+(a4-2a2c2+c4)+(b4-2b2c2+c4
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)
$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$
$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$
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)
$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$
$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$
$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$