\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)
a) \(A=x^2-3x+5\)
\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\("="\Leftrightarrow x=5\Rightarrow x=0;5\)
c) \(C=4x-x^2+3\)
\("="\Leftrightarrow x=7\Rightarrow x=2;7\)
d) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
a, A <=> \(x^2-2x\frac{3}{2}+\left(\frac{3}{2}\right)^2+2,75\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+2,75\)
ta có \(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\Rightarrow A\ge2,75\)
=> Min A =2,75 \(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)
b, \(B\Leftrightarrow4x^2-4x+1+x^2+4x+4\)
\(B\Leftrightarrow5x^2+5\)
Ta có \(5x^2\ge0\Rightarrow B\ge5\)
=> Min B = 5 <=> x=0
c,\(C\Leftrightarrow-\left(x^2-4x+4-7\right)\)
\(C\Leftrightarrow7-\left(x-2\right)^2\)
Ta có \(\left(x-2\right)^2\ge0\Rightarrow C\le7\)
=> Max C=7 <=> ( x - 2 )2 = 0 <=> x=2
d, \(C=x^4+x^2+2\)
Lại có \(x^4+x^2\ge0\)
\(\Rightarrow C\ge2\). Để Min C= 2 <=> \(x^4+x^2=0\Leftrightarrow x^2\left(x^2+1\right)=0\)\(\Leftrightarrow x=0\)
f,F \(F\Leftrightarrow-\left(\left(3x\right)^2-2.3x.2+2^2-19\right)\)
\(F\Leftrightarrow19-\left(3x-2\right)^2\)
ta có \(\left(3x-2\right)^2\ge0\)
=> \(F\le19\)
Để Max F =19 <=> x=\(\frac{2}{3}\)
