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5. 若$x^{2} + x - 1 = 0$，求$\frac{x^{4} + (x - 1)^{2} - 1}{x(x - 1)}$的值.
答案
解：$\because x^2 + x - 1 = 0$，
$\therefore x^2 = -(x - 1)$，
$\therefore \frac{x^4 + (x - 1)^2 - 1}{x(x - 1)} = \frac{x^4 + x^2 - 2x + 1 - 1}{-x^3} = \frac{x^3 + x - 2}{-x^2} = \frac{x(x^2 + 1) - 2}{-x^2} = \frac{x(-x + 1 + 1) - 2}{-x^2} = \frac{-x^2 + 2x - 2}{-x^2} = \frac{-x^2 + 2(x - 1)}{-x^2} = \frac{-x^2 - 2x^2}{-x^2} = 3$。
$\therefore x^2 = -(x - 1)$，
$\therefore \frac{x^4 + (x - 1)^2 - 1}{x(x - 1)} = \frac{x^4 + x^2 - 2x + 1 - 1}{-x^3} = \frac{x^3 + x - 2}{-x^2} = \frac{x(x^2 + 1) - 2}{-x^2} = \frac{x(-x + 1 + 1) - 2}{-x^2} = \frac{-x^2 + 2x - 2}{-x^2} = \frac{-x^2 + 2(x - 1)}{-x^2} = \frac{-x^2 - 2x^2}{-x^2} = 3$。
6. 已知$xy \neq 0$，且$\begin{cases}x + 4y - 3z = 0, \\ 4x - 5y + 2z = 0.\end{cases}$
(1)用含$z$的代数式表示$x$，$y$；
(2)求$\frac{3x^{2} + 2xy + z^{2}}{x^{2} + y^{2}}$的值.
(1)用含$z$的代数式表示$x$，$y$；
(2)求$\frac{3x^{2} + 2xy + z^{2}}{x^{2} + y^{2}}$的值.
答案
解：(1)$\begin{cases}x + 4y - 3z = 0, &①\\4x - 5y + 2z = 0, &②\end{cases}$
$①\times4 - ②$，得$21y - 14z = 0$，
$\therefore y = \frac{2}{3}z$，
把$y = \frac{2}{3}z$代入①，得$x + 4\times\frac{2}{3}z - 3z = 0$，
$\therefore x = \frac{1}{3}z$。
(2) 把$x = \frac{1}{3}z,y = \frac{2}{3}z$代入$\frac{3x^2 + 2xy + z^2}{x^2 + y^2}$，得
原式$=\frac{3\times\frac{1}{9}z^2 + 2\times\frac{1}{3}z\times\frac{2}{3}z + z^2}{\frac{1}{9}z^2 + \frac{4}{9}z^2} = \frac{16}{5}$。
$①\times4 - ②$，得$21y - 14z = 0$，
$\therefore y = \frac{2}{3}z$，
把$y = \frac{2}{3}z$代入①，得$x + 4\times\frac{2}{3}z - 3z = 0$，
$\therefore x = \frac{1}{3}z$。
(2) 把$x = \frac{1}{3}z,y = \frac{2}{3}z$代入$\frac{3x^2 + 2xy + z^2}{x^2 + y^2}$，得
原式$=\frac{3\times\frac{1}{9}z^2 + 2\times\frac{1}{3}z\times\frac{2}{3}z + z^2}{\frac{1}{9}z^2 + \frac{4}{9}z^2} = \frac{16}{5}$。
7. (1)已知$\frac{a - b}{a + b} = 3$，求代数式$\frac{5(a - b)}{a + b} - \frac{3a + 3b}{a - b}$的值；
(2)已知实数$x$满足$x + \frac{1}{x} = 4$，求分式$\frac{x}{x^{2} + 3x + 1}$的值.
(2)已知实数$x$满足$x + \frac{1}{x} = 4$，求分式$\frac{x}{x^{2} + 3x + 1}$的值.
答案
解：(1) 原式$=5\times\frac{a - b}{a + b} - 3\times\frac{a + b}{a - b}$，
$\because\frac{a - b}{a + b} = 3$，
$\therefore\frac{a + b}{a - b} = \frac{1}{3}$，
$\therefore$原式$=5\times3 - 3\times\frac{1}{3} = 15 - 1 = 14$。
(2) 原式$=\frac{1}{x + 3 + \frac{1}{x}}$，
当$x + \frac{1}{x} = 4$时，原式$=\frac{1}{4 + 3} = \frac{1}{7}$。
$\because\frac{a - b}{a + b} = 3$，
$\therefore\frac{a + b}{a - b} = \frac{1}{3}$，
$\therefore$原式$=5\times3 - 3\times\frac{1}{3} = 15 - 1 = 14$。
(2) 原式$=\frac{1}{x + 3 + \frac{1}{x}}$，
当$x + \frac{1}{x} = 4$时，原式$=\frac{1}{4 + 3} = \frac{1}{7}$。
8. (2023·鼓楼区自主招生)已知$a + b + c = 2025$，$\frac{a}{x^{2} - yz} = \frac{b}{y^{2} - xz} = \frac{c}{z^{2} - xy}$，$x + y + z \neq 0$，求$\frac{ax + by + cz}{x + y + z}$的值.
答案
解：由题意，设$\frac{a}{x^2 - yz} = \frac{b}{y^2 - xz} = \frac{c}{z^2 - xy} = k$，
$\therefore a = k(x^2 - yz),b = k(y^2 - xz),c = k(z^2 - xy)$。
$\therefore$原式$=\frac{kx(x^2 - yz) + ky(y^2 - xz) + kz(z^2 - xy)}{x + y + z} = \frac{k(x^3 + y^3 + z^3 - 3xyz)}{x + y + z} = \frac{k[(x + y)^3 + z^3 - 3xy(x + y) - 3xyz]}{x + y + z} = \frac{k(x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz)}{x + y + z} = k(x^2 - yz) + k(y^2 - xz) + k(z^2 - xy) = a + b + c = 2025$。
$\therefore a = k(x^2 - yz),b = k(y^2 - xz),c = k(z^2 - xy)$。
$\therefore$原式$=\frac{kx(x^2 - yz) + ky(y^2 - xz) + kz(z^2 - xy)}{x + y + z} = \frac{k(x^3 + y^3 + z^3 - 3xyz)}{x + y + z} = \frac{k[(x + y)^3 + z^3 - 3xy(x + y) - 3xyz]}{x + y + z} = \frac{k(x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz)}{x + y + z} = k(x^2 - yz) + k(y^2 - xz) + k(z^2 - xy) = a + b + c = 2025$。
