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八年级数学解答题一般
题目
已知关于aabb的方程组{ab=1+3ma+b=7m\left\{\begin{array}{l}a-b=1+3m\\ a+b=-7-m\end{array}\right.中,aa为负数,bb为非正数.
(1)(1)mm的取值范围;
(2)(2)mm的取值范围内,当mm为何整数时,不等式2mx+x<2m+12mx+x \lt 2m+1的解集为x>1x \gt 1.
知识点:绝对值(二)、解二元一次方程组——代入消元法、解一元一次不等式组、解二元一次方程组章节:第3章 一元一次不等式 / 3.5 一元一次不等式组

答案与解析

答案

(1){ab=1+3ma+b=7m\left\{\begin{array}{l}{a-b=1+3m①}\\{a+b=-7-m②}\end{array}\right.
((++)÷2)\div 2得:a=m3a=m-3③,
将③代入②得:3+m+b=7m-3+m+b=-7-m
解得:b=2m4b=-2m-4
\therefore方程组{ab=1+3ma+b=7m\left\{\begin{array}{l}a-b=1+3m\\ a+b=-7-m\end{array}\right.的解为{a=m3b=2m4\left\{\begin{array}{l}{a=m-3}\\{b=-2m-4}\end{array}\right..
a\because a为负数,bb为非正数,
{m302m40\therefore \left\{\begin{array}{l}{m-3<0}\\{-2m-4≤0}\end{array}\right.
解得:2m<3-2\leqslant m \lt 3
m\therefore m的取值范围为2m<3-2\leqslant m \lt 3
(2)2mx+x<2m+1(2)\because 2mx+x \lt 2m+1
(2m+1)x<2m+1\therefore \left(2m+1\right)x \lt 2m+1.
\because不等式2mx+x<2m+12mx+x \lt 2m+1的解集为x>1x \gt 1
2m+1<0\therefore 2m+1 \lt 0
m<12\therefore m \lt -\frac{1}{2}
2m<3\because -2\leqslant m \lt 3
2m<12\therefore -2\leqslant m \lt -\frac{1}{2}
m=1\therefore m=-1m=2m=-2
\thereforemm2-21-1时,不等式2mx+x<2m+12mx+x \lt 2m+1的解集为x>1x \gt 1.

解析

(1){ab=1+3ma+b=7m\left\{\begin{array}{l}{a-b=1+3m①}\\{a+b=-7-m②}\end{array}\right.
((++)÷2)\div 2得:a=m3a=m-3③,
将③代入②得:3+m+b=7m-3+m+b=-7-m
解得:b=2m4b=-2m-4
\therefore方程组{ab=1+3ma+b=7m\left\{\begin{array}{l}a-b=1+3m\\ a+b=-7-m\end{array}\right.的解为{a=m3b=2m4\left\{\begin{array}{l}{a=m-3}\\{b=-2m-4}\end{array}\right..
a\because a为负数,bb为非正数,
{m302m40\therefore \left\{\begin{array}{l}{m-3<0}\\{-2m-4≤0}\end{array}\right.
解得:2m<3-2\leqslant m \lt 3
m\therefore m的取值范围为2m<3-2\leqslant m \lt 3
(2)2mx+x<2m+1(2)\because 2mx+x \lt 2m+1
(2m+1)x<2m+1\therefore \left(2m+1\right)x \lt 2m+1.
\because不等式2mx+x<2m+12mx+x \lt 2m+1的解集为x>1x \gt 1
2m+1<0\therefore 2m+1 \lt 0
m<12\therefore m \lt -\frac{1}{2}
2m<3\because -2\leqslant m \lt 3
2m<12\therefore -2\leqslant m \lt -\frac{1}{2}
m=1\therefore m=-1m=2m=-2
\thereforemm2-21-1时,不等式2mx+x<2m+12mx+x \lt 2m+1的解集为x>1x \gt 1.

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