a) Cho \(a>0,b>0\), giải và biện luận bất phương trình:
       \(x+1>\frac{bx}{a}+\frac{a}{b}              (1)\)
b) Cho \(abc<0\), giải và biện luận phương trình          
       \(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}>2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})     (2)\)
Giải
a) Vì \(a>0,b>0\) nên \((1) \Leftrightarrow abx-b^2x>a^2-ab \Leftrightarrow (a-b)bx>(a-b)a\)
- Nếu \(a>b\) thì \(x>\frac{a}{b}\) vậy tập nghiệm \(S=(\frac{a}{b};+\infty )\)
- Nếu \(a<b\) thì \(x<\frac{a}{b}\) vậy tập nghiệm \(S=(-\infty; \frac{a}{b} )\)
- Nếu $(a=b) $ thì $ 0x=0$, đúng với mọi x nên tập nghiệm S=$R$.
b) \((2) \Leftrightarrow (\frac{x-a}{bc}-\frac{1}{b}-\frac{1}{c})+(\frac{x-b}{ca}-\frac{1}{c}-\frac{1}{a})+(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b})>0\)
            \(\Leftrightarrow (x-a-b-c)(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})>0\)
            \(\Leftrightarrow (a+b+c)(x-a-b-c)<0\) ( vì \(abc<0\))
- Nếu \(a+b+c>0\) thì \(x<a+b+c:S=(-\infty ;a+b+c)\)
- Nếu \(a+b+c<0\) thì \(x>a+b+c:S=(a+b+c;+\infty )\)
- Nếu \(a+b+c=0\) thì \(S=\varnothing \)

Thẻ

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