Giải và biện luận phương trình sau theo các  tham số :
$m(mx+n)=4mx+n^2-5$.
Ta có:
$m \neq 0$  và  $m \neq 4      \Rightarrow           x=\frac{n^2-mn-5}{m(m-4)} $.
$m=0           \Rightarrow            \begin{cases}n \neq \pm \sqrt 5     \Rightarrow  S = \emptyset \\ n=\pm\sqrt 5   \Rightarrow  S=\mathbb{R} \end{cases} $;
$m=4           \Rightarrow            0.x = n^2-4n-5=(n+1)(n-5)$
                           $\Rightarrow      \begin{cases}n\neq -1; n \neq 5                \Rightarrow      S=\emptyset \\ n=-1  \text { hoặc } n=5   \Rightarrow    S=\mathbb{R}\end{cases}$
Thax b. Mình làm đc rùi. –  lovet.o.p.1990 06-07-12 05:19 PM
Gợi ý cho bạn ;) –  Kit Nguyen 06-07-12 01:45 PM

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