1) $A=\dfrac{1}{16\sqrt[]{17}+68}+\dfrac{1}{17\sqrt[]{18}+18\sqrt[]{17}}+...+\dfrac{1}{x\sqrt[]{x+1}+(x+1)\sqrt[]{x}}=\dfrac{499}{2012}$
Ta có:
$\dfrac{1}{x\sqrt[]{x+1}+(x+1)\sqrt[]{x}}$
$=\dfrac{1}{\sqrt[]{x(x+1)}(\sqrt[]{x}+\sqrt[]{x+1})}$
$=\dfrac{\sqrt[]{x}-\sqrt[]{x+1}}{-\sqrt[]{x(x+1)}}$
$=\dfrac{1}{\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x+1}}$
$→A=\dfrac{1}{\sqrt[]{16}}-\dfrac{1}{\sqrt[]{17}}+\dfrac{1}{\sqrt[]{17}}-\dfrac{1}{\sqrt[]{18}}+...+\dfrac{1}{\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x+1}}$
$=\dfrac{1}{4}-\dfrac{1}{\sqrt[]{x+1}}=\dfrac{499}{2012}$ (Điều kiện: $x>-1$)
$↔16\sqrt[]{x+1}=8048$
$↔\sqrt[]{x+1}=503$
$↔x+1=253009$
$↔x=253008$ (thỏa mãn)
Vậy $x=253008$ là giá trị cần tìm.
2) Ta có:
$P=(xy+1)(x^3+y^3)-5(x^2+y^2)+14x^2y^2-58xy+6$
$=(xy+1)(x+y)(x^2-xy+y^2)-5(x+y)^2-48xy+14(xy)^2+6$
$=(xy+1)(x+y)[(x+y)^2-3xy]-5(x+y)^2-48xy+14(xy)^2+6$
Theo bất đẳng thức $Cô-si$, ta có:
$x+y≥2\sqrt[]{xy}$
$↔4≥2\sqrt[]{xy}$
$↔\sqrt[]{xy}≤2$
$↔0≤xy≤4$
$→ GTNN$ là:
$P_{min}=4.16-5.16+6=-10$
$→ GTLN$ là:
$P_{max}=5.4^2-5.16-48.4+14.16+6=38$.
