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13.已知$4x - y = 4a^2 - 8a + 16$，$x + y = a^2 + 8a - 16$，若$x ≤ y$，则$\frac{a^2 -1}{a -1}$的值为
5
。答案
13.5 【点拨】$\because 4x-y=4a^2-8a+16$,$x+y=a^2+8a-16$,$\therefore \begin{cases}4x-y=4a^2-8a+16,\\x+y=a^2+8a-16,\end{cases}$解得$\begin{cases}x=a^2,\\y=8a-16,\end{cases}$$\because x≤ y$,$\therefore a^2-8a+16≤0$,$\therefore (a-4)^2≤0$,$\because (a-4)^2≥0$,$\therefore a-4=0$,解得$a=4$,$\therefore \frac{a^2-1}{a-1}=\frac{(a-1)(a+1)}{a-1}=a+1=4+1=5$.
14. 关于$x$的不等式组$\begin{cases}x-\dfrac{m}{3}≥ 0, \\ 2x-3≥ 3(x-2)\end{cases}$恰有两个整数解，且$\dfrac{4(m+1)}{m^2 -1}$的值为正整数，则整数$m$的值为 ______ 。
答案
14.5 【点拨】由题意可知不等式组$\begin{cases}x-\frac{m}{3}≥0,\\2x-3≥3(x-2)\end{cases}$有解,$\therefore$解不等式组$\begin{cases}x-\frac{m}{3}≥0,\\2x-3≥3(x-2),\end{cases}$得它的解集为$\frac{m}{3}≤ x≤3$.又$\because$不等式组恰有两个整数解,$\therefore 1＜\frac{m}{3}≤2$.$\therefore 3＜m≤6$.$\therefore$整数$m$的值为4,5,6.$\because \frac{4(m+1)}{m^2-1}=\frac{4(m+1)}{(m+1)(m-1)}=\frac{4}{m-1}$,$\therefore$当$m=5$时,分式的值为正整数.$\therefore$整数$m$的值为5.
15. 已知$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$，$\frac{1}{b}+\frac{1}{c}=\frac{1}{3}$，$\frac{1}{a}+\frac{1}{c}=\frac{1}{5}$，求$\frac{abc}{ab+bc+ac}$的值。
答案
15.【解】$\because \frac{1}{a}+\frac{1}{b}=\frac{1}{2}$,$\frac{1}{b}+\frac{1}{c}=\frac{1}{3}$,$\frac{1}{a}+\frac{1}{c}=\frac{1}{5}$,
$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}$.
$\therefore 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{15+10+6}{30}=\frac{31}{30}$.$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{31}{60}$.$\therefore \frac{abc}{ab+bc+ac}=\frac{1}{\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}=\frac{60}{31}$.
$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}$.
$\therefore 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{15+10+6}{30}=\frac{31}{30}$.$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{31}{60}$.$\therefore \frac{abc}{ab+bc+ac}=\frac{1}{\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}=\frac{60}{31}$.
16. 已知$xyz≠0$，且$\begin{cases}4x - 3y - 6z = 0, \\x + 2y - 7z = 0.\end{cases}$
(1) 用含$z$的代数式表示$x,y$；
(2) 求$\dfrac{2x + 3y - 6z}{x + 5y + 7z}$的值。
(1) 用含$z$的代数式表示$x,y$；
(2) 求$\dfrac{2x + 3y - 6z}{x + 5y + 7z}$的值。
答案
16.【解】(1)$\begin{cases}4x-3y-6z=0,①\\x+2y-7z=0,②\end{cases}$
由②$×4-$①,得$11y-22z=0$,$\therefore y=2z$.
把$y=2z$代入②,得$x+4z-7z=0$,$\therefore x=3z$.
(2)$\because x=3z$,$y=2z$,$\therefore \frac{2x+3y-6z}{x+5y+7z}=\frac{6z+6z-6z}{3z+10z+7z}=\frac{6z}{20z}=\frac{3}{10}$.
由②$×4-$①,得$11y-22z=0$,$\therefore y=2z$.
把$y=2z$代入②,得$x+4z-7z=0$,$\therefore x=3z$.
(2)$\because x=3z$,$y=2z$,$\therefore \frac{2x+3y-6z}{x+5y+7z}=\frac{6z+6z-6z}{3z+10z+7z}=\frac{6z}{20z}=\frac{3}{10}$.
17. 在初中数学学习阶段，我们常常会利用一些变形技巧来简化式子，解答问题.
材料一:在解决某些分式问题时,倒数法是常用的变形技巧之一,所谓倒数法,即把式子变成其倒数形式,从而运用约分化简,以达到计算目的.
例:若$\frac{x}{x^2 + 1}=\frac{1}{4}$,求代数式$x^2 + \frac{1}{x^2}$的值.
解:易知$x≠0$. $\because \frac{x}{x^2 + 1}=\frac{1}{4}$, $\therefore \frac{x^2 +1}{x}=4$.
即$\frac{x^2}{x}+\frac{1}{x}=4$, $\therefore x+\frac{1}{x}=4$.
$\therefore x^2 + \frac{1}{x^2}=(x+\frac{1}{x})^2 -2=16-2=14$.
材料二:在解决某些连等式问题时,通常可以引入参数"k",将连等式变成几个值为k的等式,这样就可以通过适当变形解决问题.
例:若$2x=3y=4z$,且$xyz≠0$,求$\frac{x}{y+z}$的值.
解:令$2x=3y=4z=k(k≠0)$,
则$x=\frac{k}{2}, y=\frac{k}{3}, z=\frac{k}{4}$,
$\therefore \frac{x}{y+z}=\frac{\frac{1}{2}k}{\frac{1}{3}k+\frac{1}{4}k}=\frac{\frac{1}{2}}{\frac{7}{12}}=\frac{6}{7}$.
根据材料回答问题:
(1)已知$\frac{x}{x^2 -x +1}=\frac{1}{4}$,求$x^2 + \frac{1}{x^2}$的值.
(2)已知$\frac{a}{5}=\frac{b}{4}=\frac{c}{3}(abc≠0)$,求$\frac{3b +4c}{2a}$的值.
(3)已知$x,y,z$为实数,$\frac{xy}{x+y}=-2,\frac{yz}{y+z}=\frac{4}{3}$, $\frac{zx}{z+x}=-\frac{4}{3}$,求分式$\frac{xyz}{xy + yz + zx}$的值.
材料一:在解决某些分式问题时,倒数法是常用的变形技巧之一,所谓倒数法,即把式子变成其倒数形式,从而运用约分化简,以达到计算目的.
例:若$\frac{x}{x^2 + 1}=\frac{1}{4}$,求代数式$x^2 + \frac{1}{x^2}$的值.
解:易知$x≠0$. $\because \frac{x}{x^2 + 1}=\frac{1}{4}$, $\therefore \frac{x^2 +1}{x}=4$.
即$\frac{x^2}{x}+\frac{1}{x}=4$, $\therefore x+\frac{1}{x}=4$.
$\therefore x^2 + \frac{1}{x^2}=(x+\frac{1}{x})^2 -2=16-2=14$.
材料二:在解决某些连等式问题时,通常可以引入参数"k",将连等式变成几个值为k的等式,这样就可以通过适当变形解决问题.
例:若$2x=3y=4z$,且$xyz≠0$,求$\frac{x}{y+z}$的值.
解:令$2x=3y=4z=k(k≠0)$,
则$x=\frac{k}{2}, y=\frac{k}{3}, z=\frac{k}{4}$,
$\therefore \frac{x}{y+z}=\frac{\frac{1}{2}k}{\frac{1}{3}k+\frac{1}{4}k}=\frac{\frac{1}{2}}{\frac{7}{12}}=\frac{6}{7}$.
根据材料回答问题:
(1)已知$\frac{x}{x^2 -x +1}=\frac{1}{4}$,求$x^2 + \frac{1}{x^2}$的值.
(2)已知$\frac{a}{5}=\frac{b}{4}=\frac{c}{3}(abc≠0)$,求$\frac{3b +4c}{2a}$的值.
(3)已知$x,y,z$为实数,$\frac{xy}{x+y}=-2,\frac{yz}{y+z}=\frac{4}{3}$, $\frac{zx}{z+x}=-\frac{4}{3}$,求分式$\frac{xyz}{xy + yz + zx}$的值.
答案
17.【解】(1)易知$x≠0$. $\because \frac{x}{x^2 -x +1}=\frac{1}{4}$,$\therefore \frac{x^2 -x +1}{x}=4$,
$\therefore x+\frac{1}{x}=5$,$\therefore x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=23$.
(2)由$\frac{a}{5}=\frac{b}{4}=\frac{c}{3}(abc≠0)$可设$a=5k$,$b=4k$,$c=3k(k≠0)$,
$\therefore \frac{3b+4c}{2a}=\frac{3×4k+4×3k}{2×5k}=\frac{12}{5}$.
(3)$\because \frac{xy}{x+y}=-2$,$\therefore \frac{x+y}{xy}=-\frac{1}{2}$,
$\therefore \frac{1}{x}+\frac{1}{y}=-\frac{1}{2}$,同理可得$\frac{1}{y}+\frac{1}{z}=\frac{3}{4}$,$\frac{1}{x}+\frac{1}{z}=-\frac{3}{4}$,
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{z}=-\frac{1}{2}$,
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{1}{4}$,
$\therefore \frac{xy+yz+zx}{xyz}=-\frac{1}{4}$,
$\therefore \frac{xyz}{xy+yz+zx}=-4$.
$\therefore x+\frac{1}{x}=5$,$\therefore x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=23$.
(2)由$\frac{a}{5}=\frac{b}{4}=\frac{c}{3}(abc≠0)$可设$a=5k$,$b=4k$,$c=3k(k≠0)$,
$\therefore \frac{3b+4c}{2a}=\frac{3×4k+4×3k}{2×5k}=\frac{12}{5}$.
(3)$\because \frac{xy}{x+y}=-2$,$\therefore \frac{x+y}{xy}=-\frac{1}{2}$,
$\therefore \frac{1}{x}+\frac{1}{y}=-\frac{1}{2}$,同理可得$\frac{1}{y}+\frac{1}{z}=\frac{3}{4}$,$\frac{1}{x}+\frac{1}{z}=-\frac{3}{4}$,
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{z}=-\frac{1}{2}$,
$\therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{1}{4}$,
$\therefore \frac{xy+yz+zx}{xyz}=-\frac{1}{4}$,
$\therefore \frac{xyz}{xy+yz+zx}=-4$.
