2022寒兴趣三阶第3讲 一次方程和方程组的解法

  1. 方程$\left( {k}^{2}-4 \right){x}^{2}+\left( 2-3k \right)x+\left( k-2 \right)y+3k=0$为二元一次方程,则$k$的值为
    ( ).
    A. $2$
    B. $-2$
    C. $2$或$-2$
    D. 以上均不对

    【答案】$\text{B}$

    • 由题意,得
      $\begin{cases} k^2-4=0 \\ k-2\ne 0 \\2-3k\ne 0\end{cases}$,
      $\therefore k=-2$,
      故选:$\text{B}$.
  2. 若$\left( a-1 \right){x}^{\left| a \right|}+a=5$是关于$x$的一元一次方程,则$a$的值是$\underline{\hspace{3em}}$.

    【答案】$-1$

    • 由一元一次方程的定义,
      可知$\left| a \right|=1$,且$a-1\ne 0$,
      解得$a=-1$.
  3. 解下列方程:
    (1)$\dfrac{1-0.5x}{0.3}-\dfrac{0.2x-1}{0.3}=\dfrac{0.3x}{0.02}$.
    (2)$\dfrac{0.1x-0.4}{1.2}-1=\dfrac{0.2x+1}{0.3}$.
    (3)$\dfrac{7x-1}{0.024}=\dfrac{1-2x}{0.018}-\dfrac{5x+1}{0.012}$.

    【答案】(1)$x=\dfrac{5}{13}$.(2)$x=-8$.(3)$x=\dfrac{1}{59}$.

    • (1)如下
    • $\begin{align}1-0.5x-\left(0.2x-1\right)&=15\left(0.3x\right)\\
      x&=\dfrac{5}{13}\end{align}$
      (2)如下
      $\begin{align}0.1x-0.4-1.2 &=4\left(0.2x+1\right)\\
      x=-8\end{align}$
      (3)如下
      $\begin{align}3\left(7x-1\right) &=4\left(1-2x\right)-6\left(5x+1\right)\\
      x=\dfrac{1}{59}\end{align}$
  4. 解下列方程:
    (1)$1-\left\{ 1-\left[ 1-\left( 1-x \right)-x \right]-x \right\}+x=2$.
    (2)$\dfrac{1}{2}\left\{ x-\dfrac{1}{3}\left[ x-\dfrac{1}{4}\left( x-\dfrac{2}{3} \right) \right] -\dfrac{3}{2} \right\}=x+\dfrac{3}{4}$.

    【答案】(1)$x=1$;(2)$x=-\dfrac{22}{9}$.

    • (1)如下
      $\begin{align}
      1-\left[{1-\left(1-x\right)-x} \right]-x &=x-1 \\
      1-\left(1-x\right)-x &=2-2x \\
      1-x &=x-1 \\
      x &=1
      \end{align}$
      (2)如下
      $\begin{align}
      x-\dfrac{1}{3}\left[x-\dfrac{1}{4}\left(x-\dfrac{2}{3} \right) \right]-\dfrac{3}{2} &=2x+\dfrac{3}{2} \\
      \dfrac{1}{3}\left[x-\dfrac{1}{4}\left(x-\dfrac{2}{3}\right)\right] &=-x-3 \\
      x-\dfrac{1}{4}\left( x-\dfrac{2}{3} \right) &=-3x-9 \\
      \dfrac{1}{4}\left( x-\dfrac{2}{3} \right) &=4x+9 \\
      x-\dfrac{2}{3} &=16x+36 \\
      -\dfrac{110}{3} &=15x \\
      x &=-\dfrac{22}{9} \\
      \end{align}$
  5. 解方程:$\dfrac{1}{3}-\dfrac{1}{2}\left( \dfrac{3}{5}x-7 \right)=\dfrac{1}{2}-\dfrac{1}{3}\left( 7-\dfrac{3}{5}x \right)$.

    【答案】$x=\dfrac{34}{3}$

    • 这一方程在变换过程中,应将$\left( \dfrac{3}{5}x-7 \right)$视为一个整体.方程两边同时乘以$6$,得
      $2-3\left(\dfrac{3}{5}x-7\right)=3-2\left(7-\dfrac{3}{5}x\right)$,$-3\left(\dfrac{3}{5}x-7\right)+2\left(7-\dfrac{3}{5}x\right)=3-2$,$-3\left(\dfrac{3}{5}x-7\right)-2\left(\dfrac{3}{5}x-7\right)=1$,
      $-5\left(\dfrac{3}{5}x-7\right)=1$,即$\dfrac{3}{5}x=\dfrac{34}{5}$,解得$x=\dfrac{34}{3}$.
  6. 解方程:$x+\dfrac{x+1}{2}+\dfrac{x+2}{3}+\cdots +\dfrac{x+2014}{2015}+\dfrac{x+2015}{2016}=2016$.

    【答案】$ x=1$.

    • $x-1+\dfrac{x}{2}-\dfrac{1}{2}+\dfrac{x}{3}-\dfrac{1}{3}+\cdots+\dfrac{x}{2015}-\dfrac{1}{2015}+\dfrac{x}{2016}-\dfrac{1}{2016}=0$,
      $x+\dfrac{x}{2}+\dfrac{x}{3}+\cdots +\dfrac{x}{2015}+\dfrac{x}{2016}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{2015}+\dfrac{1}{2016}$,
      $ x=1$.
  7. $x+\dfrac{x}{1+2}+\dfrac{x}{1+2+3}+\cdots +\dfrac{x}{1+2+3+\cdots 2016}=2016$,求$x$的值.

    【答案】$x=\dfrac{2017}{2}$.

    • $x\left( \dfrac{2}{1\times 2}+\dfrac{2}{2\times 3}+\dfrac{2}{3\times 4}+\cdots +\dfrac{2}{2016\times 2017} \right)=2016$
      $x=\dfrac{2017}{2}$.
  8. 三个同学对问题“若方程组$\begin{cases}a_{1}x+b_{1}y=c_{1} \\ a_{2}x+b_{2}y=c_{2} \end{cases}$的解是$\begin{cases}x=3 \\ y=4 \end{cases}$,求方程组$\begin{cases}3a_{1}x+2b_{1}y=5c_{1} \\ 3a_{2}x+2b_{2}y=5c_{2} \end{cases}$的解.”提出各自的想法.甲说:“这个题目好象条件不够,不能求解”;乙说:“它们的系数有一定的规律,可以试试”;丙说:“能不能把第二个方程组的两个方程的两边都除以$5$,通过换元替换的方法来解决”.参考他们的讨论,你认为这个题目的解应该是$\underline{\hspace{3em}}$.

    【答案】$\begin{cases}x=5 \\ y=10 \end{cases}$

    • 将$\begin{cases}x=3 \\ y=4 \end{cases}$ 代入原方程组得$\begin{cases}3a_{1}+4b_{1}=c_{1} ①\\ 3a_{2}+4b_{2}=c_{2} ② \end{cases}$,
      在$\begin{cases}3a_{1}x+2b_{1}y=5c_{1} ③\\ 3a_{2}x+2b_{2}y=5c_{2} ④ \end{cases}$中将①代入③中可以整理为$3a_{1}x+2b_{1}y=c_{1}+4c_{1}=c_{1}+4\left( 3a_{1}+4b_{1} \right)$,
      即$3a_{1}x+2b_{1}y=c_{1}+12a_{1}+16b_{1}$ ,
      整理得$a_{1}\left( 3x-12 \right)+b_{1}\left( 2y-16 \right)=c_{1}$ ,
      由第一个方程组可得:$\begin{cases}3x-12=3 \\ 2y-16=4 \end{cases}$ ,
      解得$\begin{cases}x=5 \\ y=10 \end{cases}$.
      把方程组$\begin{cases}3a_{1}x+2b_{1}y=5c_{1} \\ 3a_{2}x+2b_{2}y=5c_{2} \\\end{cases}$的两个方程的两边都除以$5$,
      得$\begin{cases}\dfrac{3}{5}a_{1}x+\dfrac{2}{5}b_{1}y=c_{1} \\ \dfrac{3}{5}a_{2}x+\dfrac{2}{5}b_{2}y=c_{2} \\\end{cases}$,
      令$\begin{cases}m=\dfrac{3}{5}x \\ n=\dfrac{2}{5}y \\\end{cases}$,
      则得$\begin{cases}a_{1}m+b_{1}n=c_{1} \\ a_{2}m+b_{2}n=c_{2} \\\end{cases}$,
      由已知得,此方程组的解为$\begin{cases}m=3 \\ n=4 \\\end{cases}$,
      即$\begin{cases}\dfrac{3}{5}x=3 \\ \dfrac{2}{5}y=4 \\\end{cases}$,
      ∴$\begin{cases}x=5 \\ y=10 \\\end{cases}$.
      故答案为:$\begin{cases}x=5 \\ y=10 \\\end{cases}$.
  9. 解下列方程组:
    (1)$\begin{cases}3\left(y-1\right)=4\left(x-4\right) \\ 5\left(x-1\right)=3\left(y+5\right) \end{cases}$.
    (2)$\begin{cases}\dfrac{x+y}{2}-\dfrac{x-y}{3}=\dfrac{7}{6} \\ \dfrac{x-y}{2}-\dfrac{x+y}{3}=-\dfrac{4}{3} \end{cases}$.
    (3)$\begin{cases}\dfrac{1}{4}\left( 3x-2 \right)-\dfrac{1}{5}\left( 2y-1 \right)=2 \\ 4\left( 3x+2 \right)+\dfrac{1}{5}\left( 3y+1 \right)=0 \end{cases}$.

    【答案】(1)$\begin{cases}m=6 \\ n=12 \end{cases}$.
    $\qquad\qquad$(2)$\begin{cases}x=-\dfrac{1}{2} \\ y=\dfrac{3}{2} \end{cases}$.
    $\qquad\qquad$(3)$\begin{cases}x=-\dfrac{38}{105} \\ y=-\dfrac{45}{7} \end{cases}$.

    • (1)$\begin{cases}4m-3n=-12 \\ 3m+2n=42 \end{cases}\Rightarrow \begin{cases}m=6 \\ n=12 \end{cases}$.
      (2)$\begin{cases}3\left( x+y \right)-2\left( x-y \right)=7 \\ 3\left( x-y \right)-2\left( x+y \right)=-8 \end{cases}\Rightarrow \begin{cases}x=-\dfrac{1}{2} \\ y=\dfrac{3}{2} \end{cases}$.
      (3)$\begin{cases}5\left( 3x-2 \right)-4\left( 2y-1 \right)=40 \\ 20\left( 3x+2 \right)+\left( 3y+1 \right)=0 \end{cases}\Rightarrow \begin{cases}x=-\dfrac{38}{105} \\ y=-\dfrac{45}{7} \end{cases}$.
  10. 解方程组:$\begin{cases}2011x+2013y=4023 \\ 2013x+2011y=4025 \\\end{cases}$.

    【答案】$\begin{cases}x=1.5 \\ y=0.5 \\\end{cases}$.

    • 两方程相减,得:$x-y=1$①,
      两方程相加,得:$y+x=2$②,
      ①+②得:$x=1.5$;
      ②-①得:$y=0.5$,
      ∴方程的解为:$\begin{cases}x=1.5 \\ y=0.5 \\\end{cases}$.
  11. 解下列方程组:
    (1)$\begin{cases}x+y-z=11 \\ y+z-x=3 \\ z+x-y=1 \end{cases}$.
    (2) $\begin{cases}8x+13y+13z=154 \\ 13x+8y+13z=154 \\ 43x+36y+67z=706 \end{cases}$.
    (3)$\begin{cases}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4} \\ 5x+2y-3z=8 \end{cases}$.

    【答案】(1)$\begin{cases}x=6 \\ y=7 \\ z=2 \end{cases}$.
    $\qquad\qquad$(2)$\begin{cases}x=3 \\ y=3 \\ z=7 \end{cases}$.
    $\qquad\qquad$(3)$\begin{cases}x=4 \\ y=6 \\ z=8 \end{cases}$.

    • (1)$x+y+z=15\Rightarrow \begin{cases}x=6 \\ y=7 \\ z=2 \end{cases}$.
      (2)$x-y=0\Rightarrow \begin{cases}x=3 \\ y=3 \\ z=7 \end{cases}$.
      (3)$6x=4y=3z\Rightarrow \begin{cases}x=4 \\ y=6 \\ z=8 \end{cases}$.
  12. 解下列方程:
    (1)$\dfrac{0.1x-0.02}{0.002}-\dfrac{0.1x+0.1}{0.05}=0.3$.
    (2)$\dfrac{x-4}{30\%}-\dfrac{x+2}{50\%}=1.7$.
    (3)$\dfrac{x+3}{0.5}+\dfrac{\dfrac{1}{3}\left( x+4 \right)}{0.125}=5x+19$.

    【答案】(1)$x=\dfrac{41}{160}$.
    $\qquad\qquad$(2)$x=14.275$.
    $\qquad\qquad$(3)$x=-7$.

  13. 解下列方程组:
    (1)$\begin{cases}\dfrac{7}{9}x-\dfrac{3}{2}y=-\dfrac{37}{9} \\ 3.6x-2.3y=-5.1 \\\end{cases}$.
    (2)$\begin{cases}\dfrac{1}{x-1}+\dfrac{2}{6y-3}=1 \\ \dfrac{1}{2x-2}+\dfrac{1}{2y-1}=0 \\\end{cases}$.
    (3)$\begin{cases}3x+5\left( x+y \right)=36 \\ 3y+4\left( x+y \right)=36 \\\end{cases}$.
    (4)$\begin{cases}107x+40y=147 \\ 170x+271y=441 \\\end{cases}$.

    【答案】(1)$\begin{cases}x=\dfrac{1}{2} \\ y=3 \\\end{cases}$.
    $\qquad\qquad$(2)$\begin{cases}x=\dfrac{5}{3} \\ y=-\dfrac{1}{6} \\\end{cases}$.
    $\qquad\qquad$(3)$\begin{cases}x=2 \\ y=4 \\\end{cases}$.
    $\qquad\qquad$(4)$\begin{cases}x=1 \\ y=1 \\\end{cases}$.

    • (1)化简得$\begin{cases}14x-27y=-74①\\ 36x-23y=-51② \\\end{cases}$
      由①得$x=\dfrac{27y-74}{14}$③,
      把③代入②得$36\cdot\dfrac{27y-74}{14}-23y=-51$,
      解得$y=3$,
      ∴$x=\dfrac{27\times3-74}{14}=\dfrac{1}{2}$.
      ∴$\begin{cases}x=\dfrac{1}{2} \\ y=3 \\\end{cases}$是此方程组的解.
      (2)令$\dfrac{1}{x-1}=m$,$\dfrac{1}{2y-1}=n$,
      ∴$\begin{cases}m+\dfrac{2}{3}n=1①\\\dfrac{1}{2}m+n=0②\\\end{cases}$
      由②得$n=-\dfrac{1}{2}m$,
      ∴$m-\dfrac{1}{3}m=1$.
      ∴$m=\dfrac{3}{2}$.
      ∴$n=-\dfrac{1}{2}\times\dfrac{3}{2}=-\dfrac{3}{4}$.
      ∴$x=\dfrac{5}{3}$,$y=-\dfrac{1}{6}$.
      ∴$\begin{cases}x=\dfrac{5}{3} \\ y=-\dfrac{1}{6} \\\end{cases}$是此方程组的解.
      (3)$\begin{cases}8x+5y=36①\\4x+7y=36②\\\end{cases}$
      ②$\times2$得$8x+14y=72$③,
      ③$-$①得$9y=36$
      $y=4$,
      把$y=4$代入①得$8x+20=36$,
      ∴$x=2$.
      ∴$\begin{cases}x=2 \\ y=4 \\\end{cases}$是此方程组的解.
      (4)$\begin{cases}107x+40y=147①\\ 170x+271y=441② \\\end{cases}$
      ①$\times3$得$321x+120y=441$③,
      ③$-$②得$151x-151y=0$,
      ∴$x=y$.
      ∴$107x+40x=147$
      $x=1$.
      ∴$x=y=1$.
      ∴$\begin{cases}x=1 \\ y=1 \\\end{cases}$是此方程组的解.
  14. 解方程组:$\begin{cases}x_{1}+x_{2}=x_{2}+x_{3}=x_{3}+x_{4}=\cdots =x_{2020}+x_{2021}=1 \\ x_{1}+x_{2}+\cdots +x_{2020}+x_{2021}=2021 \\\end{cases}$.

    【答案】$x_{1}=x_{3}=\cdots =x_{2021}=1011$,$x_{2}=x_{4}=\cdots =x_{2020}=-1010$.

  15. 已知$\begin{cases}x_{1}+x_{4}+x_{6}+x_{7}=39 \\ x_{2}+x_{4}+x_{5}+x_{7}=49 \\ x_{3}+x_{5}+x_{6}+x_{7}=41 \\ x_{4}+x_{7}=13 \\ x_{5}+x_{7}=14 \\ x_{6}+x_{7}=9 \\ x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}=9 \end{cases}$,求$x_{7}$的值.

    【答案】$-84$.

    • $\begin{cases}x_{1}+x_{4}+x_{6}+x_{7}=39&① \\ x_{2}+x_{4}+x_{5}+x_{7}=49&② \\ x_{3}+x_{5}+x_{6}+x_{7}=41&③ \\ x_{4}+x_{7}=13&④ \\ x_{5}+x_{7}=14&⑤ \\ x_{6}+x_{7}=9&⑥ \\ x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}=9&⑦ \\\end{cases}$
      用①式$-$④式,则
      $x_{1}+x_{6}=39-13=26$ ⑧
      用②式$-$⑤式,则
      $x_{2}+x_{4}=49-14=35$ ⑨
      用③式$-$⑥式,则
      $x_{3}+x_{5}=41-9=32$ ⑩
      ⑧$+$⑨$+$⑩$=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=26$
      $+35+32=93$⑪
      用⑪式$-$⑦式
      $-x_{7}=93-9=84$
      ∴$x_{7}=-84$.
  16. 若$x_{1}$,$x_{2}$,$x_{3}$,$x_{4}$,$x_{5}$满足方程组:$\begin{cases}x_{1}-x_{2}+x_{3}=1 \\ x_{2}-x_{3}+x_{4}=2 \\ x_{3}-x_{4}+x_{5}=3 \\ x_{4}-x_{5}+x_{1}=4 \\ x_{5}-x_{1}+x_{2}=5 \\\end{cases}$,求$x_{2}x_{3}x_{4}$的值.

    【答案】$126$.

    • $\begin{cases}x_{1}=0 \\ x_{2}=6 \\ x_{3}=7 \\ x_{4}=3 \\ x_{5}=-1 \\\end{cases}$,$x_{2}x_{3}x_{4}=126$.
  17. 已知$\begin{cases}\dfrac{xy}{x+y}=1 \\ \dfrac{yz}{y+z}=\dfrac{1}{2} \\ \dfrac{zx}{z+x}=\dfrac{1}{4} \end{cases}$,求$x+y+z$.

    【答案】$-\dfrac{14}{15}$.

    • $\begin{cases}\dfrac{1}{x}+\dfrac{1}{y}=1 \\ \dfrac{1}{y}+\dfrac{1}{z}=2 \\ \dfrac{1}{z}+\dfrac{1}{x}=4 \end{cases}$,解得$\begin{cases}\dfrac{1}{x}=\dfrac{3}{2} \\ \dfrac{1}{y}=-\dfrac{1}{2} \\ \dfrac{1}{z}=\dfrac{5}{2} \end{cases}$;
      $x=\dfrac{2}{3}$,$y=-2$,$z=\dfrac{2}{5}$.
      则$x+y+z=-\dfrac{14}{15}$.
  18. 设$x_{1}$,$x_{2}$,$x_{3}$,$x_{4}$,$x_{5}$,$x_{6}$,$x_{7}$是自然数,且$x_{1}<{}x_{2}<{}x_{3}<{}x_{4}<{}x_{5}<{}x_{6}<{}x_{7}$,$x_{1}+x_{2}=x_{3}$,$x_{2}+x_{3}=x_{4}$,$x_{3}+x_{4}=x_{5}$,$x_{4}+x_{5}=x_{6}$,$x_{5}+x_{6}=x_{7}$,又$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}=2010$.求$x_{1}+x_{2}+x_{3}$的最大值.

    【答案】$236$.

    • 先用$x_{1}$和$x_{2}$表示$x_{3}$,$x_{4}$,$\cdots $,$x_{7}$,得
      $\begin{cases}x_{3}=x_{1}+x_{2} \\ x_{4}=x_{2}+x_{3}=x_{1}+2x_{2} \\ x_{5}=x_{3}+x_{4}=2x_{1}+3x_{2} \\ x_{6}=x_{4}+x_{5}=3x_{1}+5x_{2} \\ x_{7}=x_{5}+x_{6}=5x_{1}+8x_{2} \end{cases}$,
      因此$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}=13x_{1}+20x_{2}=2010$,
      于是得$x_{2}=\dfrac{2010-13x_{1}}{20}=100+\left( \dfrac{1}{2}-\dfrac{13}{20}x_{1} \right)$,
      因为$x_{2}$是自然数,
      所以$\dfrac{1}{2}-\dfrac{13}{20}x_{1}$是整数,
      所以$x_{1}$是$10$的奇数倍.
      又因为$x_{1}<{}x_{2}$,故有三种解:$x_{1}=10$,$x_{2}=94$或$x_{1}=30$,$x_{2}=81$或$x_{1}=50$,$x_{2}=68$,
      所以$x_{1}+x_{2}+x_{3}$的最大值$=2\left(x_{1}+x_{2}\right)$的最大值$=236$.
© 版权声明
THE END
喜欢就支持一下吧
点赞13 分享
评论 抢沙发
头像
欢迎做题
提交
头像

昵称

取消
昵称表情代码图片

    暂无评论内容