6. 按图示的程序计算,若开始输入的 $ x $ 为正整数,最后输出的结果为 $ 40 $,则 $ x $ 的值是______。

答案
解:情况一:直接输出
$3x + 1 = 40$
$3x = 39$
$x = 13$
情况二:一次循环后输出
$3(3x + 1) + 1 = 40$
$9x + 3 + 1 = 40$
$9x = 36$
$x = 4$
情况三:两次循环后输出
$3[3(3x + 1) + 1] + 1 = 40$
$27x + 13 = 40$
$27x = 27$
$x = 1$
情况四:三次循环后输出
$3[3(3(3x + 1) + 1) + 1] + 1 = 40$
$81x + 40 = 40$
$x = 0$(非正整数,舍去)
综上,$x$的值是$1$,$4$,$13$
答案:$1$,$4$,$13$
$3x + 1 = 40$
$3x = 39$
$x = 13$
情况二:一次循环后输出
$3(3x + 1) + 1 = 40$
$9x + 3 + 1 = 40$
$9x = 36$
$x = 4$
情况三:两次循环后输出
$3[3(3x + 1) + 1] + 1 = 40$
$27x + 13 = 40$
$27x = 27$
$x = 1$
情况四:三次循环后输出
$3[3(3(3x + 1) + 1) + 1] + 1 = 40$
$81x + 40 = 40$
$x = 0$(非正整数,舍去)
综上,$x$的值是$1$,$4$,$13$
答案:$1$,$4$,$13$
7. 解下列方程:
(1) $ 3x + 7 = 32 - 2x $;
(2) $ \frac{3y - 1}{4} - 1 = \frac{5y - 7}{6} $;
(3) $ 3(20 - y) = 6y - 4(y - 11) $;
(4) $ 5(x + \frac{1}{2}) - 9 = 7(x + \frac{1}{2}) - 13 $。
(1) $ 3x + 7 = 32 - 2x $;
(2) $ \frac{3y - 1}{4} - 1 = \frac{5y - 7}{6} $;
(3) $ 3(20 - y) = 6y - 4(y - 11) $;
(4) $ 5(x + \frac{1}{2}) - 9 = 7(x + \frac{1}{2}) - 13 $。
答案
(1)解:$3x + 2x = 32 - 7$
$5x = 25$
$x = 5$
(2)解:$3(3y - 1) - 12 = 2(5y - 7)$
$9y - 3 - 12 = 10y - 14$
$9y - 10y = -14 + 15$
$-y = 1$
$y = -1$
(3)解:$60 - 3y = 6y - 4y + 44$
$60 - 3y = 2y + 44$
$-3y - 2y = 44 - 60$
$-5y = -16$
$y = \frac{16}{5}$
(4)解:$5(x + \frac{1}{2}) - 7(x + \frac{1}{2}) = -13 + 9$
$-2(x + \frac{1}{2}) = -4$
$x + \frac{1}{2} = 2$
$x = \frac{3}{2}$
$5x = 25$
$x = 5$
(2)解:$3(3y - 1) - 12 = 2(5y - 7)$
$9y - 3 - 12 = 10y - 14$
$9y - 10y = -14 + 15$
$-y = 1$
$y = -1$
(3)解:$60 - 3y = 6y - 4y + 44$
$60 - 3y = 2y + 44$
$-3y - 2y = 44 - 60$
$-5y = -16$
$y = \frac{16}{5}$
(4)解:$5(x + \frac{1}{2}) - 7(x + \frac{1}{2}) = -13 + 9$
$-2(x + \frac{1}{2}) = -4$
$x + \frac{1}{2} = 2$
$x = \frac{3}{2}$
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