\(Q=2x^2-6x\)

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

\(=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge\frac{-9}{2}\forall x\)

Dấu"=" xảy ra<=>\(2\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

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

\(=x^2+y^2+5x+10\)

\(=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+10+y^2\)

\(=\left(x+\frac{5}{2}\right)^2+y^2+\frac{15}{4}\)

Vì \(\hept{\begin{cases}\left(x+\frac{5}{2}\right)^2\ge0;\forall x,y\\y^2\ge0;\forall x,y\end{cases}}\)

\(\Rightarrow\left(x+\frac{5}{2}\right)^2+y^2\ge0;\forall x,y\)

\(\Rightarrow\left(x+\frac{5}{2}\right)^2+y^2+\frac{15}{4}\ge0+\frac{15}{4};\forall x,y\)

Hay \(M\ge\frac{15}{4};\forall x,y\)

Dấu =" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{5}{2}\right)^2=0\\y^2=0\end{cases}}\)

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

Vậy MIN \(M=\frac{15}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{-5}{2}\\y=0\end{cases}}\)

các anh chị pro toán giúp em

có ai giúp em không

a) Đặt \(A=10+2x-5x^2\)

\(-A=5x^2-2x-10\)

\(-5A=25x^2-10x-50\)

\(-5A=\left(25x^2-10x+1\right)-51\)

\(-5A=\left(5x-1\right)^2-51\)

Do \(\left(5x-1\right)^2\ge0\forall x\)

\(\Rightarrow-5A\ge-51\)

\(A\le\frac{51}{5}\)

Dấu "=" xảy ra khi : \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy Max A = \(\frac{51}{5}\Leftrightarrow x=\frac{1}{5}\)

b) Đặt \(B=x^2-6x+10\)

\(B=\left(x^2-6x+9\right)+1\)

\(B=\left(x-3\right)^2+1\)

Mà \(\left(x-3\right)^2\ge0\forall x\)

\(B\ge1\)

Dấu "=" xảy ra khi :

\(x-3=0\Leftrightarrow x=3\)

Vậy Min B \(=1\Leftrightarrow x=3\)

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

2x² + y² + 2xy - 6x - 2y + 10

= x² + y² + 1² + 2xy - 2x - 2y + x² - 4x + 4 + 5

= (x + y - 1)² + (x - 2)² + 5  5

Dấu ''='' xảy ra khi {x+y−1=0x−2=0{x+y−1=0x−2=0 ⇔{y=−1x=2⇔{y=−1x=2

Vậy Min = 5 khi x = 2 và y = - 1

Ta có: \(B=2x^2+y^2-2xy+6x+10\)

\(=x^2-2xy+y^2+x^2+6x+9+1\)

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

Dấu '=' xảy ra khi x=y=-3

Vậy: \(B_{min}=1\) khi (x,y)=(-3;-3)

1. x²-8x+1 = x² -2x.4 + 4² - 4² +1 = ( x- 4 )² - 15 
mà ( x - 4 )²  > 0
=> ( x - 4 )² -15 > 0

Vậy -15 là gt min của biểu thức khi x = 4

2. x² - 4x + y² - 6y + 2 = x² - 2.2x + 2² + y² - 2.3y + 3² -11 = (x-2)² + ( y - 3)² -11
mà ( x - 2)² > 0
      ( y - 3)² > 0 
Vậy -11 là gt min của biểu thức khi x=2 và y = 3

Mình nghĩ là bài 3 là tìm gt lớn nhất chứ bạn ^^

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)

Ta có : 2x² - 6x 

= \(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.6+36-36\)

Q\(=\left(\sqrt{2}x-6\right)^2-36\)

Vì \(\left(\sqrt{2}x-6\right)^2\ge0\forall x\)

Nên : Q = \(=\left(\sqrt{2}x-6\right)^2-36\) \(\ge-36\forall x\)

Vậy \(Q_{min}=-36\) khi \(\sqrt{2}x-6=0\) => \(\sqrt{2}x=6\) => \(x=6:\sqrt{2}=3\sqrt{2}\)