2 Tìm giá trị nhỏ nhất của biểu thức :
a) A=2x^2-8x+10 ; b) B= x(x+1)(x^2+x-4)
d) D=9x^2-6x+5 ; e) E=x^2+3-1
f) F=(x^2+5x+4)(x+2)(x+3) ; G= (x-3)(x-4)(x^2-7x+8)
h) H=\(\left|x-7\right|+\left|x+5\right|\) ; i) I=\((2x-1)^2-3\left|2x-1\right|+2\)
k) K=\(\left|x^2+x+1\right|+\left|x^2+x-12\right|\)
h) Ta có: \(\left\{{}\begin{matrix}\left|x-7\right|=\left|7-x\right|\ge7-x\\\left|x+5\right|\ge x+5\end{matrix}\right.\)
\(\Rightarrow\left|7-x\right|+\left|x+5\right|\ge\left(7-x\right)+\left(x+5\right)\)
\(\Rightarrow\left|x-7\right|+\left|x+5\right|\ge12\)
\(\Rightarrow H\ge12\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}7-x\ge0\\x+5\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le7\\x\ge-5\end{matrix}\right.\)
\(\Leftrightarrow-5\le x\le7\)
Vậy, MinH = 12 \(\Leftrightarrow-5\le x\le7\)
a) Ta có: \(A=2x^2-8x+10\)
\(=2\left(x^2-4x+5\right)\)
\(=2\left(x^2-4x+2^2+1\right)\)
\(2\left[\left(x-2\right)^2+1\right]\)
Ta lại có: \(\left(x-2\right)^2\ge0\)
\(\Rightarrow2\left[\left(x-2\right)^2+1\right]\ge2\)
\(\Rightarrow A\ge2\)
Dấu bằng xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy MinA = 2 \(\Leftrightarrow x=2\)