\(\Delta'=\left(m+1\right)^2-\left(2m+10\right)=m^2-9\)

Pt có 2 nghiệm khi \(m^2-9\ge0\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le-3\end{matrix}\right.\)

Khi đó theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m+10\end{matrix}\right.\)

\(A=x_1^2+x_2^2+14x_1x_2=\left(x_1+x_2\right)^2+12x_1x_2\)

\(=4\left(m+1\right)^2+12\left(2m+10\right)\)

\(=4\left(m+4\right)^2+60\ge60\)

Dấu "=" xảy ra khi \(m+4=0\Rightarrow m=-4\) (thỏa mãn)

a.

\(y'=x^2+2\left(m^2-1\right)x+2m-3\)

\(y''=2x+2\left(m^2-1\right)\)

Hàm đạt cực đại tại \(x=2\) khi: \(\left\{{}\begin{matrix}y'\left(2\right)=0\\y''\left(2\right)< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4+4\left(m^2-1\right)+2m-3=0\\4+2\left(m^2-1\right)< 0\end{matrix}\right.\)

Do \(2m^2+2>0\) ;\(\forall m\) nên ko tồn tại m thỏa mãn yêu cầu đề bài

b.

\(y'=x^2+2mx+3\)

\(y''=2x+2m\)

Hàm đạt cực đại tại \(x=-3\) khi: \(\left\{{}\begin{matrix}9-6m+3=0\\-6+2m< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m=2\\m< 3\end{matrix}\right.\)

\(\Rightarrow m=2\)

\(P=\dfrac{1}{\left(x+1\right)^2+5}\le\dfrac{1}{5}\)

\(P_{max}=\dfrac{1}{5}\) khi \(x+1=0\Rightarrow x=-1\)

\(Q=\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{4x^2+4x+4}{4\left(x+1\right)^2}=\dfrac{3\left(x^2+2x+1\right)+x^2-2x+1}{4\left(x+1\right)^2}=\dfrac{3}{4}+\dfrac{\left(x-1\right)^2}{4\left(x+1\right)^2}\)

\(Q_{min}=\dfrac{3}{4}\) khi \(x-1=0\Rightarrow x=1\)

1: \(x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5>=5\forall x\)

=>\(P=\dfrac{1}{x^2+2x+6}< =\dfrac{1}{5}\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne-\dfrac{1}{m}\)

=>\(-m^2\ne2\)(luôn đúng)

\(\left\{{}\begin{matrix}mx-y=m^2\\2x+my=m^2+2m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-m^2\\2x+m\left(mx-m^2\right)=m^2+2m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-m^2\\2x+m^2x=m^3+m^2+2m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=mx-m^2\\x\left(m^2+2\right)=\left(m+1\right)\left(m^2+2\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=mx-m^2=m\left(m+1\right)-m^2=m\end{matrix}\right.\)

Đặt \(A=x^2+3y+4\)

\(=\left(m+1\right)^2+3m+4\)

\(=m^2+5m+5\)

\(=m^2+2\cdot m\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}\)

\(=\left(m+\dfrac{5}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu '=' xảy ra khi m=-5/2

Vì \(\dfrac{2}{1}\ne\dfrac{-1}{1}=-1\)

nên hệ luôn có nghiệm duy nhất

\(\left\{{}\begin{matrix}2x-y=3m-7\\x+y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=3m-7+1=3m-6\\x+y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m-2\\y=1-m+2=-m+3\end{matrix}\right.\)

Để x,y dương thì \(\left\{{}\begin{matrix}m-2>0\\-m+3>0\end{matrix}\right.\)

=>2<m<3

\(P=x-y-xy-2m\)

\(=m-2-\left(-m+3\right)-\left(m-2\right)\left(-m+3\right)-2m\)

\(=m-2+m-3+\left(m-2\right)\left(m-3\right)-2m\)

\(=m^2-5m+6-5=m^2-5m+1\)

\(=m^2-5m+\dfrac{25}{4}-\dfrac{21}{4}=\left(m-\dfrac{5}{2}\right)^2-\dfrac{21}{4}>=-\dfrac{21}{4}\forall m\)

Dấu '=' xảy ra khi m=5/2(nhận)

-ĐKXĐ: \(x\ne5\)

\(\dfrac{\left(m^2+1\right)x+1-2m^2}{x-5}=2m\)

\(\Leftrightarrow\dfrac{\left(m^2+1\right)x+1-2m^2}{x-5}=\dfrac{2m\left(x-5\right)}{x-5}\)

\(\Rightarrow m^2x+x+1-2m^2=2mx-10m\)

\(\Leftrightarrow m^2x+x-2mx=2m^2-10m-1\)

\(\Leftrightarrow x\left(m^2-2m+1\right)=2m^2-10m-1\)

\(\Leftrightarrow x=\dfrac{2m^2-10m-1}{\left(m-1\right)^2}\)

-Để phương trình có nghiệm duy nhất, đạt GT duy nhất thì \(\left(m-1\right)^2\ne0\Leftrightarrow m\ne1\)

-Chết mình nhầm, bài đầu tiên đúng rồi nhé. Mình xin lỗi bạn!

Áp dụng BĐT Cauchy cho 2 số dương:

\(x+3\ge2\sqrt{3x}\)

\(\Rightarrow\dfrac{x+3}{\sqrt{x}}\ge\dfrac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)

\(ĐTXR\Leftrightarrow x=3\)

\(-\dfrac{3}{1-2m}=\left|\dfrac{4-5m}{1-2m}\right|\Leftrightarrow\dfrac{3}{2m-1}=\left|\dfrac{4-5m}{1-2m}\right|\)

TH1 : \(\dfrac{3}{2m-1}=\dfrac{4-5m}{1-2m}\Leftrightarrow3=5m-4\Leftrightarrow m=\dfrac{7}{5}\)(tm)

TH2 : \(\dfrac{3}{2m-1}=\dfrac{5m-4}{1-2m}\Leftrightarrow3=4-5m\Leftrightarrow m=\dfrac{1}{5}\)(tm)