Question

cho a,b thỏa a>0 và a+b>=1

tìm MIN A=\(\dfrac{8a^2+b}{4a}+b^2\)

b) tìm x,y nguyên thỏa \(x^4-2x^3+6x^2-4y^2-32x+4y+39=0\)

Answer 1

bài 1:

vì \(a+b\ge1\Leftrightarrow b\ge1-a\)

khi đó \(A\ge\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2=2a+\dfrac{1}{4a}-\dfrac{1}{4}+1-2a+a^2\)

\(=a^2+\dfrac{1}{4a}+\dfrac{3}{4}=a^2+\dfrac{1}{8a}+\dfrac{1}{8a}+\dfrac{3}{4}\)

Áp dụng BĐT cauchy:\(a^2+\dfrac{1}{8a}+\dfrac{1}{8a}\ge3\sqrt[3]{a^2.\dfrac{1}{8a}.\dfrac{1}{8a}}=\dfrac{3}{4}\)

\(\Rightarrow A\ge\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)

Dấu = xảy ra khi \(a^2=\dfrac{1}{8a}\Leftrightarrow a=\dfrac{1}{2}\Rightarrow b=\dfrac{1}{2}\)

Vậy A_MIN=\(\dfrac{3}{2}\)khi \(a=b=\dfrac{1}{2}\)

Answer 2

\(Pt\Leftrightarrow x^4-2x^3+6x^2-32x+40=\left(2y-1\right)^2\)

\(\Leftrightarrow\left(x^2+2x+10\right)\left(x-2\right)^2=\left(2y-1\right)^2\)

cách of thím thế này hả