\(M=x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}+y^2+2.3.y+9-9+10\)

\(M=\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\left(y^2+2.3.y+9\right)+\frac{3}{4}\)

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

Vậy \(M_{min}=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

Chọn mình nha cảm ơn chúc bạn học tốt

a)\(A=\left(x-5\right)^2\ge0\)

\(\Rightarrow Min=0\)dấu \(=\)xảy ra khi \(x=5\)

a) \(A=x^2-10x+25\)

\(A=\left(x^2-10x+25\right)+0\)

\(A=\left(x-5\right)^2+0\)

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

\(\Rightarrow A\ge0\)

Dấu "=" xảy ra khi :  \(x-5=0\Leftrightarrow x=5\)

Vậy ...

b) \(B=x^2+y^2-x+6y+10\)

\(B=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

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

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

       \(\left(y+3\right)^2\ge0\forall y\)

\(\Rightarrow B\ge\frac{3}{4}\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

Vậy ...

\(M=x^2+y^2-x+6y+10\)

\(M=x^2-2.\frac{1}{2}x+\frac{1}{4}+y^2+6y+9+1-\frac{1}{4}\)

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

\(M_{min}=1-\frac{1}{4}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2},y=-3\)

P/s tham khảo nha

\(x^2+y^2-x+6y+10\)

=\(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}+y^2+6y+9+\frac{3}{4}\)

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

Có \(\left(x-\frac{1}{2}\right)^2\ge0\)

       \(\left(y+3\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

                                   \(y+3=0\Rightarrow y=-3\)

Vậy MinM = \(\frac{3}{4}\)\(\Leftrightarrow\)\(x=\frac{1}{2}\)và \(y=-3\)

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

Ta có : P = x² - 2x + 5 = x² - 2x + 1 + 4 = (x - 1)² + 4

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

Suy ra : \(P=\left(x-1\right)^2+4\ge4\forall x\)

Nên : P_min = 4 khi x = 1

b) Ta có Q = 2x² - 6x = 2(x² - 3x) = 2(x² - 3x + \(\frac{9}{4}-\frac{9}{4}\) ) = \(2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)

Vì \(2\left(x-\frac{3}{2}\right)^2\ge0\forall x\) 

SUy ra ; \(Q=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

Vậy \(Q_{min}=-\frac{9}{2}\) khi \(x=\frac{3}{2}\)

a) Ta có: Q = 2x² - 6x = 2x² - 6x + 9/2 - 9/2 = 2(x² - 3x + 9/4) - 9/2 = 2(x - 3/2)² - 9/2

Ta luôn có : (x - 3/2)² \(\ge\)0 \(\forall\)x --> 2(x - 3/2)² \(\ge\)0 \(\forall\)x

     => 2(x - 3/2)² - 9/2 \(\ge\)-9/2 \(\forall\)x

hay Q \(\ge\)-9/2 \(\forall\)x

Dấu "=" xảy ra <=> (x - 3/2)² = 0 <=> x - 3/2 = 0 <=> x = 3/2

Vậy Q_min = -9/2 tại x = 3/2

b) Ta có:

M = x² + y² - x + 6y + 10 = (x² - x + 1/4) + (y² + 6y + 9) + 3/4 = (x - 1/2)² + (y + 3)² + 3/4

Ta luôn có: (x - 1/2)² \(\ge\)0 \(\forall\)x

                (y + 3)² \(\ge\) 0 \(\forall\)y

 => (x - 1/2)² + (y + 3)² + 3/4 \(\ge\) 3/4 \(\forall\)x,y 

hay M \(\ge\)3/4 \(\forall\)x , y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\) <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

Vậy M_min = 3/4 tại x = 1/2 và y = -3

cho mk hỏi chữ A ngược là sao vậy