Question

Cho biểu thức \(A=\dfrac{\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)}{x^2y^2+1+\left(x^2-y\right)\left(1-y\right)}\)

a) CMR: Biểu thức A không phụ thuộc vào biến \(x\) ?

b) Tìm Min A ?

Answer 1

a: \(B=\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)\)

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

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

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

\(=\left(x^2+1\right)\left(y+\dfrac{1}{2}\right)^2\)

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

\(=x^2y^2+1+x^2-x^2y-y+y^2\)

\(=x^2y^2-y+x^2+y^2-x^2y+1\)

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

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

=>\(A=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)

b: \(=\dfrac{y^2-y+1+2y-\dfrac{3}{4}}{y^2-y+1}=1+\dfrac{2y-\dfrac{3}{4}}{y^2-y+1}>=1\)

Dấu = xảy ra khi y=3/8