rút gon:\(B=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
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25 tháng 11 2019
trả lời:
\(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\frac{\left(x-y\right)^3+z^3+3x^2y-3xy^2+3xyz}{x^2+2xy+y^2+y^2+2yz+z^2+z^2-2xz+x^2}\)
\(=\frac{\left(x-y+z\right)\left[\left(x-y\right)^2-\left(x-y\right).z+z^2\right]+3xy\left(x-y+z\right)}{2x^2+2y^2+2z^2+2xy+2yz-2zx}\)
\(=\frac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{2\left(x^2+y^2+z^2+xy+yz-zx\right)}\)
\(=\frac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-zx\right)}{2\left(x^2+y^2+z^2+xy+yz-zx\right)}\)
\(=\frac{x-y+x}{2}\)
~hok tốt~
15 tháng 9 2023
a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)
=a+b+c
b:
Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\dfrac{x-y+z}{2}\)
ND
Nguyễn Đức Trí
VIP
15 tháng 9 2023
a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)
\(=a+b+c\)
23 tháng 9 2017
\(\dfrac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\dfrac{x\left(x+y\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{xy^2+2xyz+x^2z}.\dfrac{x^2+xy-xy-xy^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\\ =\dfrac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2xy^2+4xyz+2x^2z}.\dfrac{x^2-xy^2}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2-xy\right)}{2xy^2+4xy+2x^2z}\)
@@ ko ra nữa
Ta có : \(B=\dfrac{\left(x+z\right)^3-y^3-3xz\left(x+z\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+z^2+2xz+yz+xy+y^2\right)-3xz\left(x-y+z\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\dfrac{1}{2}\left(x-y+z\right)2\left(x^2+y^2+z^2+xy+zy-xz\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\dfrac{1}{2}\left(x-y+z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{x-y+z}{2}\)