Answer:

\(D=m^2-4mp+5p^2+10m-22p+20\)

\(=m^2-4mp+4p^2+p^2+10m-20p-2p+1+19\)

\(=\left(m^2-4mp+4p^2\right)+\left(10m-20p\right)+\left(p^2-2p+1\right)+19\)

\(=\left(m-2p\right)^2+10\left(m-2p\right)+\left(p-1\right)^2+25-6\)

\(=[\left(m-2p\right)^2+10\left(m-2p\right)+25]+\left(p-1\right)^2-6\)

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

\(\forall m;p\) có \(\left(m-2p+5\right)^2+\left(p-1\right)^2-6\ge-6\) hay \(D\ge-6\)

Dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(m-2p+5\right)^2=0\\\left(p-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}m-2p+5=0\\p-1=0\end{cases}}\Rightarrow\hept{\begin{cases}m-2p+5=0\\p=1\end{cases}}\Rightarrow\hept{\begin{cases}m-2.1+5=0\\p=1\end{cases}}\Rightarrow\hept{\begin{cases}m=-3\\p=1\end{cases}}\)

Vậy giá trị nhỏ nhất của biểu thức \(D=-6\) khi \(\hept{\begin{cases}m=-3\\p=1\end{cases}}\)

a ) \(P\left(x\right)=3x^2-27x+54=3\left(x^2-9x+15\right)\)

\(=3\left[\left(x^2-3x\right)-\left(6x-18\right)\right]=3\left[x\left(x-3\right)-6\left(x-3\right)\right].\)

\(\Rightarrow P\left(x\right)=3\left(x-3\right)\left(x-6\right)\)

Ta có : \(P\left(x\right)\ge0\Leftrightarrow\left(x-3\right)\left(x-6\right)\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x-6\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x-6\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge6\\x\le3\end{matrix}\right.\)

Vậy \(P\left(x\right)\ge0\Leftrightarrow x\le3\) hoặc \(x\ge6\)

b ) \(A=m^2-4mp+5p^2+10m-22p+28\)

\(=m^2-4mp+4p^2+10m-20p+p^2-2p+1+27\)

\(=\left(m-2p\right)^2+10\left(m-2p\right)+\left(p-1\right)^2+25+2\)

\(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)

Vậy GTNN của A là 2 khi và chỉ khi \(\left\{{}\begin{matrix}p-1=0\\m-2p+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}p=1\\m=-3\end{matrix}\right..\)

Vậy ...............

\(=3\left[\left(x^2-3x\right)-\left(6x-18\right)\right]=3\left[x\left(x-3\right)-6\left(x-3\right)\right]\)

Bài 1:

a)\(F=x^2+26y^2-10xy+14x-76y+59\)

         \(=\left(x^2-2\cdot x\cdot5y+25y^2\right)+\left(14x-70y\right)+\left(y^2-6x+9\right)+50\)

        \(=[\left(x-5y\right)^2+14\left(x-5y\right)+49]+\left(y-3\right)^2+1\)

          \(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\)

 Để Fmin=1 thì y=3;x=8

b)\(H=m^2-4mp+5p^2+10m-22p+28\)

         \(=\left(m^2-2\cdot m\cdot2p+4p^2\right)+\left(10m-20p\right)+\left(p^2-2p+1\right)+27\)

         \(=[\left(m-2p\right)^2+2\cdot\left(m-2p\right)\cdot5+25]+\left(p-1\right)^2+2\)

           \(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)

Để Hmin=2 thì p=1;m=-3

lỗi r bn ơi

Bạn ghi lại đề đi bạn

A = \(\left(m^2-4mp+4p^2\right)+10\left(m-2p\right)+25+\left(p^2-2p+1\right)+2\)

  \(=\left(m-2p\right)^2+2.5.\left(m-2p\right)+5^2+\left(p-1\right)^2+2\)

  \(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)

Vậy: A min = 2 \(\Leftrightarrow m=-3;p=1\)