\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)

Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)

Dấu \("="\Leftrightarrow x=-5\)

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

\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)

\(B=\dfrac{7}{-\left(x-5\right)^2-5}\ge-\dfrac{7}{5}\)

\(B_{min}=-\dfrac{7}{5}\) khi \(x=5\)

\(M=-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\le0\)

maxM=0 khi x=5

có max thôi không có min

Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

\(-A=2x^2+y^2+2xy-3x-2y-2\)

\(-A=\left(x^2+2xy+y^2\right)+x^2-4x-2y-2\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)-4\)

\(-A=\left(x+y-1\right)^2+\left(x-1\right)^2-4\)

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

\(B=\left(x-2y+5\right)^2+\left(y-1\right)^2+1\)

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

       \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

Các bạn giúp mình vs, mình đang cần gấp

Ta có : \(P=\frac{x^2-10x+22}{\left(x-3\right)^2}\)

Đặt : \(x-3=y\Leftrightarrow x=y+3\)

\(P=\frac{\left(y+3\right)^2-10\left(y+3\right)+22}{y^2}\)

\(P=\frac{y^2+6y+9-10y-30+22}{y^2}\)

\(P=\frac{y^2-4y+1}{y^2}\)

\(P=\frac{y^2}{y^2}-\frac{4y}{y^2}+\frac{1}{y^2}\)

\(P=1-\frac{4}{y}+\frac{1}{y^2}\)

\(P=\left(\frac{1}{y^2}-\frac{4}{y}+4\right)-3\)

\(P=\left(\frac{1}{y}-2\right)^2-3\)

Mà \(\left(\frac{1}{y}-2\right)^2\ge0\forall y\)

\(\Rightarrow P\ge-3\)

Dấu "=" xảy ra khi : 

\(\frac{1}{y}-2=0\Leftrightarrow\frac{1}{y}=2\Leftrightarrow y=\frac{1}{2}\) 

Lại có : \(x=y+3\)

\(\Rightarrow x=\frac{7}{2}\)

Vậy \(P_{Min}=-3\Leftrightarrow x=\frac{7}{2}\)

D=(x² - 4xy + 4y²) +(y² - 22y + 121) - 93

= (x-2y)² + (y-11)² - 93

Vì (x-2y)² và (y-11)² luôn lớn hơn 0 nên GTNN của biểu thức là -93

Khi đó y=11

và x=22

tách hđt #@