\(A=\left(a^4-2a^3+a^2\right)+2\left(a^2-2a+1\right)+3\)

\(A=\left(a^2-a\right)^2+2\left(a-1\right)^2+3\ge3\)

\(A_{min}=3\) khi \(a=1\)

A - 2√a nhé,  mk nhầm

a thuộc  [1;1] là sao bạn? 

\(a,x=2\Leftrightarrow A=3\cdot4-4\cdot2-1=12-8-1=3\\ b,B=x^3-1-2x+x^2-2+x-x^3=x^2-x-3\\ c,C=B-A=x^2-x-3-3x^2+3x+1=-2x^2-2x-2\\ C=-2\left(x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\right)=-2\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{2}\le-\dfrac{3}{2}\\ C_{max}=-\dfrac{3}{2}\Leftrightarrow x=-\dfrac{1}{2}\)

tach 14-x = 10-4-x roi sau do chac ban cung phai tu biet lam

\(A=5-8x+x^2=-8x+x^2+6-11\)

\(=\left(x-4\right)^2-11\)

Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy Amin = - 11 <=> x = 4

\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)

\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy Bmax = 9 <=> x = - 1

\(\left\{{}\begin{matrix}a^2\ge0\\a^4+a^2+1>0\end{matrix}\right.\) ;\(\forall a\Rightarrow P=\dfrac{a^2}{a^4+a^2+1}\ge0\)

\(P_{min}=0\) khi \(a=0\)

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

\(P_{max}=\dfrac{1}{3}\) khi \(a^2=1\Rightarrow a=\pm1\)

Ta có  \(3P=\dfrac{3a^2}{a^4+a^2+1}=\dfrac{-a^4+2a^2-1+a^4+a^2+1}{a^4+a^2+1}=1-\dfrac{\left(a^2-1\right)^2}{a^4+a^2+1}\le1\)\(\Rightarrow P\le\dfrac{1}{3}\)

Dấu "=" xảy ra <=> a² - 1 = 0 <=> a = \(\pm1\)

Vậy Max P = 1/3 khi a = \(\pm1\)

+) Dễ thấy \(P=\dfrac{a^2}{a^4+a^2+1}\ge0\) ("=" khi a = 0) 

Vậy \(0\le P\le\dfrac{1}{3}\)