Question

Chứng minh rằng: 3x² y⁴ - 5xy³ -3/2 x²y⁴ + 3xy³ + 2xy³ +1, luôn dương với mọi x,y help me!!!!!!!!!

Answer 1

\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0

Answer 2

3x² y⁴ - 5xy³ -3/2 x2y⁴ + 3xy³ + 2xy³ +1

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

Ta có:x²\(\ge0\forall x\in Q\)

y^{4\(\ge0\forall x\in Q\)}

\(\Rightarrow x^2y^4\)\(\ge0\forall x\in Q\)

\(\Rightarrow\dfrac{3}{2}\)\(x^2y^4\) \(\ge0\forall x\in Q\)

\(\Rightarrow\dfrac{3}{2}\)\(x^2y^4\) \(\ge1\forall x\in Q\) (đpcm)