Question

tìm GTNN của

a, \(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

b, \(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)

c, \(C=\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}\)

d, \(D=\dfrac{x^6+512}{x^2+8}\)

Answer 1

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)

=\(\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)

= \(\dfrac{3\cdot\left(x^2-2x+5\right)}{x^2-2x+5}+\dfrac{2}{x^2-2x+5}\)

= \(3+\dfrac{2}{x^2-2x+5}\)

= \(3+\dfrac{2}{x^2-2x+1+4}\)

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

vì (x-1)² ≥ 0 ∀ x

⇔ (x-1)² +4 ≥ 4

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

⇔\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)

⇔ A \(\le\dfrac{7}{2}\)

⇔ Min A =\(\dfrac{7}{2}\)

khi x-1=0

⇔ x=1

vậy ....

Answer 2

Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)

\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)

\(B=2-\dfrac{3}{x^2-8x+16+6}\)

\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)

\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)

Answer 3

d)\(D=\dfrac{x^6+512}{x^2+8}\)

\(D=\dfrac{x^6+8x^4-8x^4-64x^2+64x^2+512}{x^2+8}\)

\(D=\dfrac{x^4\left(x^2+8\right)-8x^2\left(x^2+8\right)+64\left(x^2+8\right)}{x^2+8}\)

\(D=\dfrac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)

\(D=x^4-8x^2+64\)

\(D=\left(x^2-4\right)^2+48\ge48\)

\(\Rightarrow MIND=48\Leftrightarrow x=\pm2\)

Answer 4

\(C=\dfrac{x^6+27}{x^4-3x^3+6x^2-9x+9}\\ \\ =\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4-3x^3+3x^2+3x^2-9x+9}\\ =\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{\left(x^4-3x^3+3x^2\right)+\left(3x^2-9x+9\right)}\\ =\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^2\left(x^2-3x+3\right)+3\left(x^2-3x+3\right)}\\ =\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ =\dfrac{x^4-3x^2+9}{x^2-3x+3}\\ =\dfrac{x^4-9x^2+3x^2+3x^2+9-3x^3+3x^3+9x-9x}{x^2-3x+3}\\ =\dfrac{\left(x^4-3x^3+3x^2\right)+\left(3x^3-9x^2+9x\right)+\left(3x^2-9x+9\right)}{x^2-3x+3}\\ =\dfrac{x^2\left(x^2-3x+3\right)+3x\left(x^2-3x+3\right)+3\left(x^2-3x+3\right)}{x^2-3x+3}\\ =\dfrac{\left(x^2+3x+3\right)\left(x^2-3x+3\right)}{x^2-3x+3}\\ =x^2+3x+3\)

\(=x^2+3x+\dfrac{9}{4}+\dfrac{3}{4}\\ =\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}\\ =\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\)

Do \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow C=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra khi:

\(\left(x+\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{3}{2}=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

Vậy \(C_{\left(Min\right)}=\dfrac{3}{4}\) khi \(x=-\dfrac{3}{2}\)