Question

\(24x^2-20x+5=0\)

Answer 1

\(24x^2-20x+5=0\)

\(\Leftrightarrow x^2-\dfrac{5}{6}x+\dfrac{5}{24}=0\)

\(\Leftrightarrow x^2-2\cdot\dfrac{5}{12}\cdot x+\left(\dfrac{5}{12}\right)^2+\dfrac{5}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{5}{12}\right)^2+\dfrac{5}{144}=0\)

=> PTVN

Answer 2

Ta có : \(24x^2-20x+5=24\left(x^2-\dfrac{20}{24}x+\dfrac{5}{24}\right)=24\left[\left(x^2-2.\dfrac{20}{48}.x+\dfrac{25}{144}+\dfrac{5}{144}\right)\right]=24\left[\left(x-\dfrac{20}{48}\right)^2+\dfrac{5}{144}\right]\ge\dfrac{24.5}{144}=\dfrac{5}{6}>0\left(\forall x\right)\)

Do đó không có giá trị nào của x thỏa mãn.