Question

Chứng minh rằng với mọi x, ta có:

a) \(\frac{15}{4x^2-12x+19}\le\frac{3}{2}\)?

b) \(\frac{4x+3}{x^2+1}\le4\)?

Answer 1

a) Để \(\frac{15}{4x^2-12x+19}\le\frac{3}{2}\) thì \(15\cdot2\le3\cdot\left(4x^2-12x+19\right)\)

\(\Leftrightarrow30\le12x^2-36x+57\)

\(\Leftrightarrow30-12x^2+36x-57\le0\)

\(\Leftrightarrow-12x^2+36x-27\le0\)

\(\Leftrightarrow-12\left(x^2-3x+\frac{9}{4}\right)\le0\)

\(\Leftrightarrow-12\left(x-\frac{3}{2}\right)^2\le0\)(luôn đúng)

b) Để \(\frac{4x+3}{x^2+1}\le4\)

thì \(4x+3\le4\left(x^2+1\right)\)

\(\Leftrightarrow4x+3\le4x^2+4\)

\(\Leftrightarrow4x+3-4x^2-4\le0\)

\(\Leftrightarrow-4x^2+4x-1\le0\)

\(\Leftrightarrow-\left(4x^2-4x+1\right)\le0\)

\(\Leftrightarrow-\left(2x-1\right)^2\le0\)(luôn đúng)