đk : x khác 2; x khác 3; x khác 1
\(a.A=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right)\cdot\frac{x^2-4x+3}{x^4+x^2+1}\)
\(A=\left(\frac{x^2}{\left(x-2\right)\left(x-3\right)}+\frac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(A=\left(\frac{x^2\left(x-1\right)+x^2\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\right)\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(A=\frac{x^2\left(x-1+x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(A=\frac{x^2\left(2x-4\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\frac{2x^2}{x^4+x^2+1}\)
\(b.\frac{1}{A}=\frac{x^4+x^2+1}{2x^2}=\frac{x^2}{2}+\frac{1}{2}+\frac{1}{2x^2}\) (x khác 0)
\(\frac{1}{A}=\frac{2x^2}{4}+\frac{1}{2}+\frac{1}{2x^2}\)
có 2x^2/4 và 1/2x^2 > 0 áp dụng bđt cô si ta có
\(\frac{2x^2}{4}+\frac{1}{2x^2}\ge2\sqrt{\frac{2x^2}{4}\cdot\frac{1}{2x^2}}=1\)
\(\Rightarrow\frac{1}{A}\ge\frac{3}{2}\)
\(\Rightarrow A\le\frac{2}{3}\)
DẤU = xảy ra khi 2x^2/4 = 1/2x^2 => 4x^4 = 4
=> x^4 = 1
=> x = 1 (loại) hoặc x = -1 (thỏa mãn)
vậy max a = 2/3 khi x = -1
