\(N1=\dfrac{m^2+2m-8}{m^2-2m+1}\) \(do:m\ne1\Rightarrow m^2-2m+1=\left(m-1\right)^2>0\)
\(\Rightarrow N\left(m^2-2m+1\right)=m^2+2m-8\Leftrightarrow Nm^2-2mN+N=m^2+2m-8\)
\(\Leftrightarrow\left(N-1\right)m^2-2m\left(N+1\right)+N+8=0\)
\(TH:N=1\Rightarrow m=2,25\left(tm\right)\)
\(TH:N\ne1\Rightarrow\Delta'\ge0\Leftrightarrow\left(N+1\right)^2-\left(N+8\right)\left(N-1\right)\ge0\Leftrightarrow N\le\dfrac{9}{5}\Leftrightarrow m=3,5\)
\(từ:2TH\Rightarrow maxN=\dfrac{9}{5}\Leftrightarrow m=3,5\)
(mấy ý kia làm tương tự)
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