第85页 - 江苏版启东中学作业本八年级数学（上下册）

解：∵$x^2+x-1=0，$∴$x^2=-(x-1)$

∴$\frac {x^4+(x-1)^2-1}{x(x-1)}=\frac {x^4+x^2-2x+1-1}{-x^3}=\frac {x^3+x-2}{-x^2}$

$=\frac {x(x^2+1)-2}{-x^2}=\frac {x(-x+1+1)-2}{-x^2}=\frac {-x^2+2x-2}{-x^2}$

$=\frac {-x^2+2(x-1)}{-x^2}=\frac {-x^2-2x^2}{-x^2}=3$

①

②

解：$(1)①×4-②$得$21y-14z=0，$∴$y=\frac 23z$

将$y=\frac 23z$代入$①$得$x+4×\frac 23z-3z=0$

∴$x=\frac 13z$

$(2)$把$x=\frac 13z，$$y=\frac 23z$代入$\frac {3x^2+2xy+z^2}{x^2+y^2}$

得原式$=\frac {3×\frac 19z^2+2×\frac 13z×\frac 23z+z^2}{\frac 19z^2+\frac 49z^2}=\frac {16}5$

解：$(1)$原式$=5×\frac {a-b}{a+b}-3×\frac {a+b}{a-b}$

∵$\frac {a-b}{a+b}=3，$∴$\frac {a+b}{a-b}=\frac 13$

∴原式$=5×3-3×\frac 13=15-1=14$

$(2)$原式$=\frac {1}{x+3+\frac {1}{x}}$

当$x+\frac {1}{x}=4$时，原式$=\frac {1}{4+3}=\frac {1}{7}$

解：由题意，$ \frac {a}{x^2-yz}=\frac {b}{y^2-xz}=\frac {c}{z^2-xy}=k$

∴$a=k(x^2-yz)，$$b=k(y^2-xz)，$$c=k(z^2-xy)$

∴原式$=\frac {kx(x^2-yz)+ky(y^2-xz)+kz(z^2-xy)}{x+y+z}=\frac {k(x^3+y^3+z^3-3xyz)}{x+y+z}$

$=\frac {k[(x+y)^3+z^3-3xy(x+y)-3xyz]}{x+y+z}=\frac {k(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{x+y+z}$

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

$=a+b+c=2025$