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2025-02-07 08:39:17
+) Nếu x ≤-1 thì (2x^3)>y^3>(2x-1)^3 (loại)
+) Nếu x ≥1 thì (2x^3)<y^3<(2x+1)^3 (loại)
+) Nếu x = 0 thì y =0 thỏa mãn
2024-12-15 10:14:23
2024-12-15 10:09:40
\(Tacó:1=2\sqrt{ab}+\sqrt{\dfrac{a}{3}}\le\left(a+b\right)+\dfrac{1}{2}\left(\dfrac{1}{3}+b\right)=\dfrac{3a+2b}{2}+\dfrac{1}{6}\Rightarrow3a+2b\ge\dfrac{5}{3}\\ \)\(P=\dfrac{3a}{3b}+\dfrac{a}{3b}+\dfrac{b}{3b}+\dfrac{2b}{3a}+9ab+6ab=\left(\dfrac{3a}{3b}+9ab\right)+\left(\dfrac{a}{3b}+\dfrac{b}{3a}\right)+\left(\dfrac{2b}{3a}+6ab\right)\ge6a+\dfrac{2}{3}+4b\ge2\left(3a+2b\right)+\dfrac{2}{3}=4\)\(Pmin=4\Leftrightarrow a=b=\dfrac{1}{3}\)