Ta có:
\(A=x^2-4x+5\)
\(A=x^2-4x+4+1\)
\(A=\left(x-2\right)^2+1\)
Do \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2+1\ge1\)
Dấu "=" xảy ra:
\(\left(x-2\right)^2+1=1\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x=2\)
Vậy: ....
_________
\(B=x^2+3x+1\)
\(B=x^2+2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}-\dfrac{5}{4}\)
\(B=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\)
Mà: \(\left(x+\dfrac{3}{2}\right)^2\ge0\Leftrightarrow B=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
Dấu "=" xảy ra khi:
\(\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}=-\dfrac{5}{4}\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy: ....
C và D tương tự nhé
\(E=2x^2+y^2-2xy+4x+5\)
\(E=\left(x^2+4x+4\right)+\left(y^2-2xy+x^2\right)+1\)
\(E=\left(x+2\right)^2+\left(y-x\right)^2+1\)
Mà: \(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-x\right)^2\ge0\end{matrix}\right.\Leftrightarrow E=\left(x+2\right)^2+\left(y-x\right)^2+1\ge1\)
Dấu "=" xảy ra: \(x=y=-2\)
Vậy: ...
\(C=3x^2-x+5\)
\(C=3\left(x^2-\dfrac{1}{3}x+\dfrac{5}{3}\right)\)
\(C=3\left(x^2-2\cdot\dfrac{1}{6}\cdot x+\dfrac{1}{36}+\dfrac{59}{36}\right)\)
\(C=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{59}{12}\)
Tới đây c làm tiếp nha
_____________
\(D=5x^2-3x+1\)
\(D=5\cdot\left(x^2-\dfrac{3}{5}x+\dfrac{1}{5}\right)\)
\(D=5\cdot\left(x^2-2\cdot\dfrac{3}{10}\cdot x+\dfrac{9}{100}+\dfrac{11}{100}\right)\)
\(D=5\cdot\left(x-\dfrac{3}{10}\right)^2+\dfrac{11}{20}\)
Đến đây c làm tiếp nha
