1.
A =\(2x^2-8x+10=\left(x^2-2x+1\right)+\left(x^2-6x+9\right)\)
\(=\left(x-1\right)^2+\left(x-3\right)^2=\left(x-1\right)^2+\left(3-x\right)^2\)
Có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(3-x\right)^2\ge0\end{matrix}\right.\forall x\)
<=> \(\left|x-1\right|+\left|x-3\right|\)
Áp dụng bđt |a| + |b| \(\ge\) |a + b| có:
\(\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)
đẳng thức xảy ra khi \(1\le x\le3\)
Vậy ................
1.
a)
\(A=2x^2-8x+10=2\left(x^2-4x+4\right)+2\ge=2\left(x-2\right)^2+2\ge2\)
Đẳng thức xảy ra \(\Leftrightarrow x=2\)
b)
\(B=3x^2-x+20=3\left(x^2-\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{239}{12}=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{239}{12}\ge\dfrac{239}{12}\)
Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{1}{6}\)
c) ĐK: \(x\ne-1\)
\(C=\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{4x^2+4x+4}{4x^2+8x+4}\)
\(=\dfrac{3x^2+6x+3}{4x^2+8x+4}+\dfrac{x^2-2x+1}{4x^2+8x+4}\)
\(=\dfrac{3\left(x^2+2x+1\right)}{4\left(x^2+2x+1\right)}+\dfrac{\left(x-1\right)^2}{4x^2+8x+4}=\dfrac{3}{4}+\dfrac{\left(x-1\right)^2}{4x^2+8x+4}\ge\dfrac{3}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
2.
\(A=-2x^2+3x+1=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{17}{8}=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)
Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)
\(B=-5x^2+4x-19=-5\left(x^2-\dfrac{4}{5}x+\dfrac{4}{25}\right)-\dfrac{91}{5}=-5\left(x-\dfrac{2}{5}\right)^2-\dfrac{91}{5}\le\dfrac{-91}{5}\)
Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{2}{5}\)
\(C=\dfrac{3}{4x^2-4x+5}=\dfrac{3}{\left(4x^2-4x+1\right)+4}=\dfrac{3}{\left(2x-1\right)^2+4}\le\dfrac{3}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
