222222222+4≥4>0
⇒A>0(đpcm)
kick nha mình cần điểm hỏi đáp :((
\(Bài 1 : \)
\(A = x^2 + 3x + 3 \)
\(A = x^2 + 2 . x . 3 / 2 + ( 3/2)^2 - ( 3/2)^2+3\)
\(A= ( x + 3/2 )^2 + 3/4\)\(\ge\)\(3/4\)
\(Dấu " = " xảy \) \(ra \) \(\Leftrightarrow\)\(x + 3/2 = 0\)
\(\Leftrightarrow\)\(x = - 3 / 2\)
\(Min A = 3/ 4 \) \(\Leftrightarrow\) \(x = - 3 / 2\)
\(1.\)
\(A=x^2+3x+3\)
\(=x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{3}{4}\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\ge0\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}\)
\(\Rightarrow A_{min}=\frac{3}{4}\)
Dấu "=" xảy ra: \(\left(x+\frac{3}{2}\right)^2=0\)
\(x+\frac{3}{2}=0\)
\(x=-\frac{3}{2}\)
Vậy \(A\)nhỏ nhất khi \(x=-\frac{3}{2}\)
Bài 2 :
\(Ta \) \(có : \) 
\(x^2 + 5y^2 - 4xy +2x-10y+14\)
\(= [( x^2 - 4xy + 4y^2 ) + ( 2x - 4y ) + 1]\) \(+ ( y^2 - 6y + 9 ) + 4\)
\(= [( x - 2y )^2 + 2. ( x - 2y ) + 1] + ( y- 3 )^2\)
\(+ 4\)
\(= ( x - 2y + 1 )^2 + ( y - 3 )^2 + 4\)
\(Vì \) \(( x - 2y + 1 )^2\)\(\ge\)\(0\)
\(( y - 3 )^2 \)\(\ge\)\(0\)
\(4 >0\)
\(Nên \) \(( x - 2y + 1 )^2 + ( y - 6 )^2 + 4 >0\)
\(Vậy : x^2 + 5y^2 - 4xy + 2x - 10y + 14 > 0\)
