1, A = x^2 + 6x + 2018
= x^2 + 2.x.3 + 3^2 - 3^2 + 2018
= (x + 3)^2 -3^2 + 2018
= (x + 3)^2 + 2009
=>. GTNN of A là 2009
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\(A=x^2+6x+2018\)
\(A=\left(x^2+6x+9\right)+2009\)
\(A=\left(x+3\right)^2+2009\)
Mà \(\left(x+3\right)^2\ge0\forall x\)
\(\Rightarrow A\ge2009\)
Dấu "=" xảy ra khi : \(x+3=0\Leftrightarrow x=-3\)
Vậy ...
\(B=x^2-5x+20\)
\(B=\left(x^2-5x+\frac{25}{4}\right)+\frac{55}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2+\frac{55}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow B\ge\frac{55}{4}\)
Dấu "=" xảy ra khi : \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy ...
\(C=x^2+5x+10\)
\(C=\left(x^2+5x+\frac{25}{4}\right)+\frac{15}{4}\)
\(C=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\)
Mà \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow C\ge\frac{15}{4}\)
Dấu "=" xảy ra khi : \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)
Vậy ...
\(D=x^2+10x-30\)
\(D=\left(x^2+10x+25\right)-55\)
\(D=\left(x+5\right)^2-55\)
Mà \(\left(x+5\right)^2\ge0\forall x\)
\(\Rightarrow D\ge-55\)
Dấu "=" xảy ra khi : \(x+5=0\Leftrightarrow x=-5\)
Vậy ...
\(E=x^2-20x+100\)
\(E=\left(x^2-20x+100\right)\)
\(E=\left(x-10\right)^2\)
Mà \(\left(x-10\right)^2\ge0\forall x\)
\(\Rightarrow E\ge0\)
Dấu "=" xảy ra khi : \(x-10=0\Leftrightarrow x=10\)
Vậy ...
\(P=9x^2+y^2+6xy+2000\)
\(P=\left(9x^2+6xy+y^2\right)+2000\)
\(P=\left(3x+y\right)^2+2000\)
Mà \(\left(3x+y\right)^2\ge0\forall x;y\)
\(\Rightarrow P\ge2000\)
Dấu "=" xảy ra khi : \(3x=-y\)
Vậy ...
\(Q=x^2-x+5\)
\(Q=\left(x^2-x+\frac{1}{4}\right)+\frac{19}{4}\)
\(Q=\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\)
Mà \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow Q\ge\frac{19}{4}\)
Dấu "=" xảy ra khi : \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy ...
