\(A=\dfrac{2x^2}{x^4+x^2+1}=\dfrac{6x^2}{3\left(x^4+x^2+1\right)}=\dfrac{2\left(x^4+x^2+1\right)-2x^4+4x^2-2}{3\left(x^4+x^2+1\right)}\)

\(A=\dfrac{2}{3}-\dfrac{2\left(x^2-1\right)^2}{3\left(x^4+x^2+1\right)}\le\dfrac{2}{3}\)

\(A_{max}=\dfrac{2}{3}\) khi \(x^2=1\)

\(A=-\left(x^4-2x^3+3x^2-4x-2018\right)=-\left[\left(x^4+x^2+4-2x^3+4x^2-4x\right)-2x^2\right]+2022\)

\(=-\left[\left(\left(x^2\right)^2+\left(x\right)^2+\left(2\right)^2-2\cdot x^2\cdot x+2\cdot x^2\cdot2-2\cdot x\cdot2\right)-2x^2\right]+2022\)

\(=-\left[\left(x^2-x+2\right)^2-2x^2\right]+2022\le2022\)

Mong bạn thông cảm, mình không chắc là đã giải đúng, có gì bỏ qua cho mình nhé!

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(A=4-x^2+2x\)

\(=-\left(x^2-2x+1\right)+5\)

\(=-\left(x-1\right)^2+5\le5\)

Dấu "= " ra khi x=1

a) ta có A = (2x-1)²+ ( x+2)= 4x²- 4x +1 +x+2= 4x² -3x +3 = 4x²-2*2x* \(\frac{3}{4}\)+ \(\frac{9}{16}\)+ \(\frac{39}{16}\)

= (2x-\(\frac{3}{4}\))²+ \(\frac{39}{16}\)

=> (2x-\(\frac{3}{4}\))²>=0

=> A >= \(\frac{39}{16}\)

dấu = sảy ra khi x=\(\frac{3}{2}\)

vậy A_{(min) = \(\frac{39}{16}\) khi x=\(\frac{3}{2}\)}

b) lm tương tự B_(min)= -\(\frac{25}{4}\) khi x= \(\frac{5}{2}\)

c) đặt dấu trừ ra ngoài vậy C_(max)=0 khi x=2

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