C\(=-1892+2x^2+y^2-2xy+10x\)

\(=\left(x-y\right)^2+\left(x+5\right)^2-1917\ge-1917\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+5\right)^2=0\end{cases}}\Rightarrow x=y=-5\)

Vậy min C=-1917 khi x=y=-5

Đề bài thiếu dữ kiện bạn ơi sao chỉ có ẩn x ko vậy ??????????

\(B=2x^2+10x=2x^2+10x+\frac{25}{2}-\frac{25}{2}=2\left(x^2+5x+\frac{25}{4}\right)-\frac{25}{2}\)

\(=2\left(x^2+2\cdot\frac{5}{2}x+\left(\frac{5}{2}\right)^2\right)-\frac{25}{2}=2\left(x+\frac{5}{2}\right)^2-\frac{25}{2}\)

vì \(2\left(x+\frac{5}{2}\right)^2>=0;-\frac{25}{2}=-\frac{25}{2}\Rightarrow2\left(x+\frac{5}{2}\right)^2-\frac{25}{2}>=-\frac{25}{2}\)

dấu = xảy ra khi \(2\left(x+\frac{5}{2}\right)^2=0\Rightarrow x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)

vậy min của B là \(-\frac{25}{2}\)tại x=\(-\frac{5}{2}\)

\(=2\left(x^2-5x+\frac{1}{2}\right)=2\left(x^2-5x+\frac{25}{4}-\frac{23}{4}\right)=2\left[\left(x-\frac{5}{2}\right)^2-\frac{23}{4}\right]=2\left(x-\frac{5}{2}\right)^2-\frac{23}{2}\)

Suy ra giá trị nhỏ nhất của biểu thức trên là -23/2 khi x=5/2

\(P=2x^2-10x+13=2\left(x^2-5x+\frac{25}{4}\right)+\frac{1}{2}\)

                                            \(=2\left(x-\frac{5}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu "="  xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy \(P_{min}=\frac{1}{2}\Leftrightarrow x=\frac{5}{2}\)

\(P=2x^2-10x+13\)

\(P=2\left(x^2-5x+\frac{13}{2}\right)\)

\(P=2\left[\left(x^2-2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{1}{4}\right]\)

\(P=2\left[\left(x-\frac{5}{2}\right)^2-\frac{1}{4}\right]\)

\(P=2\left(x-\frac{5}{2}\right)^2-\frac{1}{2}\ge\frac{1}{2}\)

\(\Rightarrow Pmin=\frac{-1}{2}\Leftrightarrow x=\frac{5}{2}\)

tách hđt #@

ta có \(A=x^2+y^2+9-2xy-6x+6y+x^2-4x+4+2004\)

             \(=\left(x-y-3\right)^2+\left(x-2\right)^2+2004\)

vì \(\left(x-y-3\right)^2+\left(x-2\right)^2\ge0\)

=> \(A\ge2004\)

dấu = xảy ra <=> x=2 và y=-1

\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(F=2x^2-10x+20=2\left(x-\frac{5}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2},\forall x\)

\(\Rightarrow minF=\frac{15}{2}\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

F = 2x² - 10x + 20

= 2( x² - 5x + 25/4 ) + 15/2

= 2( x - 5/2 )² + 15/2 ≥ 15/2 ∀ x

Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2

=> MinF = 15/2 <=> x = 5/2

\(đk:x^2+2x+2\ne0\Leftrightarrow x^2+2x+1+1=\left(x+1\right)^2+1\ne0\left(luôn-đúng\right)\)

\(A=\dfrac{x^2+10x+16}{x^2+2x+2}\Leftrightarrow A\left(x^2+2x+2\right)=x^2+10x+16\)

\(\Leftrightarrow Ax^2+2Ax+2A-x^2-10x-16=0\)

\(\Leftrightarrow x^2\left(A-1\right)+x\left(2A-10\right)+2A-16=0\)

\(\Rightarrow\Delta\ge0\Leftrightarrow\left(2A-10\right)^2-4\left(A-1\right)\left(2A-16\right)\ge0\)

\(\Leftrightarrow4A^2-40A+100-4\left(2A^2-18A+16\right)\ge0\)

\(\Leftrightarrow-4A^2+32A+36\ge0\Rightarrow-1\le A\le9\Rightarrow\left\{{}\begin{matrix}MinA=-1\\MaxA=9\end{matrix}\right.\)

\(tại\) \(MinA=-1\) \(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=-3\)

\(tại\) \(MaxA=9\) \(dấu"='\) \(xảy\) \(ra\Leftrightarrow x=-0,5\)

camon

\(A=2x^2+4y^2+4xy+10x+12y+18\)

\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)

\(A=\left(x+2y\right)^2+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)

\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)

Do : \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)

\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)

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

Vậy \(A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)

Chúc bạn học tốt !!!