\(A=x^2+2y^2+2xy+2x-4y+2020\)

      \(=\left(x^2+y^2+1+2x+2xy+2y\right)+\left(y^2-6y+9\right)+2010\)

        \(=\left(x+y+1\right)^2+\left(y-3\right)^2+2010\ge2010\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}y=3\\x+y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=3\\x=-4\end{cases}}}\)

Vậy \(Min_A=2010\Leftrightarrow\hept{\begin{cases}x=-4\\y=3\end{cases}}\)

Chúc bạn học tốt !!!

Tham khảo :

\(A=x^2+2y^2+2xy+2x-4y+2020\)

\(=\left(x^2+y^2+1+2x+2xy+2y\right)+\left(y^2-6y+9\right)+2010\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2010\ge2010\)

Dấu ''=''= xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}x=-4\\y=3\end{cases}}\)

\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

       \(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)

         \(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

           \(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)

               \(=\frac{x+y-z}{x-y+z}\)

Ta thay : \(x=0;y=2009;z=2010\) ta được :

\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)

Chúc bạn học tốt !!!

\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)

Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :

\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)

a) \(4x^2+12x+10=\left(2x+3\right)^2+1\ge1\)

Dấu "="\(\Leftrightarrow x=-2\)

b) \(B=\left(3x-1\right)^2+4\ge4\)

Dấu "="\(\Leftrightarrow x=\frac{1}{3}\)

a, \(A=4x^2+12x+10\)

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

Dấu"=" xảy ra<=> \(\left(2x+1\right)^2=0\)

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

\(b,B=9x^2-6x+5\)

      \(=\left(3x-1\right)^2+4\ge4\forall x\)

Dấu"=" xảy ra<=> \(\left(3x-1\right)^2=0\)

                   \(\Leftrightarrow x=\frac{1}{3}\)

a) \(x^2-5x+10\)

\(=x^2-2.\frac{5}{2}x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

b) \(2x^2+8x+10\)

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

\(=2\left[\left(x+2\right)^2+1\right]\ge2>0\)

c) thay x = 1 vào thì đề sai

d) \(\left(x+5\right)\left(x-3\right)+20=x^2+2x+5\)

\(=\left(x+1\right)^2+4\ge0>0\)

\(A=x^2+y^2-4x-2y+12\)

\(=\left(x^2-4x+4\right)+\left(y^2-2x+1\right)+7\)

\(=\left(x-2\right)^2+\left(y-1\right)^2+7\ge7\)

Vậy \(A_{min}=7\Leftrightarrow\hept{\begin{cases}x-2=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

\(\left(x-3\right)\left(x^2-6x+9-x^2-3x-9\right)+9x^2+12x+39=0\)

\(\Leftrightarrow-9x\left(x-3\right)+9x^2+12x+39=0\)

\(\Leftrightarrow39x+39=0\Rightarrow x=-1\)

Câu a bạn tính ra rồi giải nha

Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết

a) \(x^3-4x^2-12x+27\)

\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)

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

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

b) \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x^2-4\right)\left(x-3\right)\)

\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)

a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)

b) \(6x-9-x^2=-\left(x-3\right)^2\)