Question

17．三個(gè)有理數(shù)a，b，c，滿足abc＞0，求$\frac{a}{|\begin{array}{l}{a}\\{\;}\end{array}|}+\frac{|\begin{array}{l}\\{\;}\end{array}|}+\frac{c}{|\begin{array}{l}{c}\\{\;}\end{array}|}$的值．

Answer 1

分析 根據(jù)題意，由于abc＞0，依據(jù)有理數(shù)乘法法則分類(lèi)討論即可．

解答 解：由abc＞0可得出：
（1）a為負(fù)，則b、c異號(hào)，負(fù)因數(shù)有2個(gè)，$\frac{a}{|\begin{array}{l}{a}\\{\;}\end{array}|}+\frac{|\begin{array}{l}\\{\;}\end{array}|}+\frac{c}{|\begin{array}{l}{c}\\{\;}\end{array}|}$=-1-1+1=-1；
（2）b為負(fù)，則a、c異號(hào)，負(fù)因數(shù)有2個(gè)，$\frac{a}{|\begin{array}{l}{a}\\{\;}\end{array}|}+\frac{|\begin{array}{l}\\{\;}\end{array}|}+\frac{c}{|\begin{array}{l}{c}\\{\;}\end{array}|}$=-1-1+1=-1；
（3）c為負(fù)，則a、b異號(hào)，負(fù)因數(shù)有2個(gè)，$\frac{a}{|\begin{array}{l}{a}\\{\;}\end{array}|}+\frac{|\begin{array}{l}\\{\;}\end{array}|}+\frac{c}{|\begin{array}{l}{c}\\{\;}\end{array}|}$=-1-1+1=-1；
（4）c為正，則a、b為正，負(fù)因數(shù)有0個(gè)，$\frac{a}{|\begin{array}{l}{a}\\{\;}\end{array}|}+\frac{|\begin{array}{l}\\{\;}\end{array}|}+\frac{c}{|\begin{array}{l}{c}\\{\;}\end{array}|}$=1+1+1=3；
所以$\frac{a}{|\begin{array}{l}{a}\\{\;}\end{array}|}+\frac{|\begin{array}{l}\\{\;}\end{array}|}+\frac{c}{|\begin{array}{l}{c}\\{\;}\end{array}|}$的值是-1或3．

點(diǎn)評(píng) 本題考查了有理數(shù)的乘法，幾個(gè)不等于零的數(shù)相乘，積的符號(hào)由負(fù)因數(shù)的個(gè)數(shù)決定：當(dāng)負(fù)因數(shù)有奇數(shù)個(gè)數(shù)，積為負(fù)；當(dāng)負(fù)因數(shù)的個(gè)數(shù)為偶數(shù)個(gè)時(shí)，積為正．