rút gọn phân số sau: $\frac{x^3+y^3+z^3-3xyz}{(x-y)^2+(y-z)^2+(z-x)^2}$
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Đáp án:
\(\dfrac{1}{2}\left(x+y+z\right)\)
Giải thích các bước giải:
\(\dfrac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\)
\(=\dfrac{\left(x+y\right)^3-3x^2y-3xy^2-3xyz+z^3}{x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2}\)
\(=\dfrac{\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+z^2-xz-yz+z^2-3xy\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
\(=\dfrac{1}{2}\left(x+y+z\right)\)
Xét tử số:
$x^3+y^3+z^3-3xyz$
$=x^3+3x^2+3xy^2+y^3+z^3-3xyz-3x^2y-3xy^2$
$=(x^3+3x^2y+3xy^2+y^3)+z^3-(3xyz+3x^2y+3xy^2)$
$=(x+y)^3+z^3-3xy(z+x+y)$
$=(x+y+z)[(x-y)^2-z(x+y)+z^2]-3xy(x+y+z)$
$=(x+y+z)(x^2+2xy+y^2-xz-yz+z^2-3xy)$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$
Xét mẫu số:
$(x-y)^2+(y-z)^2+(z-x)^2$
$=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2$
$=2x^2+2y^2+2z^2-2xy-2xz-2yz$
$=2(x^2+y^2+z^2-xy-yz-xz)$
Vậy: `\frac{x^3+y^3+z^2-3xyz}{(x-y)^2+(y-z)^2+(z-x)^2}=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{2(x^2+y^2+z^2-xy-yz-xz)}=(x+y+z)/2`