Question

cho x,y,z là 3 số nguyên dương thỏa mãn

\(\frac{1}{\sqrt{2x-3}}+\frac{4}{\sqrt{y-2}}+\frac{16}{\sqrt{3z-1}}+\sqrt{2x-3}+\sqrt{y-2}+\sqrt{3z-1}=14\)

vậy a=......,b=......,c=........

Answer 1

Áp dụng bđt AM-GM cho 2 số dương ta có:

\(\left(\frac{1}{\sqrt{2x-3}}+\sqrt{2x-3}\right)+\left(\frac{4}{\sqrt{y-2}}+\sqrt{y-2}\right)\)\(+\left(\frac{16}{\sqrt{3z-1}}+\sqrt{3z-1}\right)\ge\)\(2\sqrt{\frac{1}{\sqrt{2x-3}}.\sqrt{2x-3}}+2\sqrt{\frac{4}{\sqrt{y-2}}.\sqrt{y-2}}\)\(+2\sqrt{\frac{16}{\sqrt{3z-1}}.\sqrt{3z-1}}=2.1+2.2+2.4=14\)

Dau "=" xay ra khi \(\left\{\begin{matrix}\frac{1}{\sqrt{2x-3}}=\sqrt{2x-3}\\\frac{4}{\sqrt{y-2}}=\sqrt{y-2}\\\frac{16}{\sqrt{3z-1}}=\sqrt{3z-1}\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}2x-3=1\\y-2=4\\3z-1=16\end{matrix}\right.\)=> \(\left\{\begin{matrix}x=1\\y=6\\z=\frac{17}{3}\end{matrix}\right.\) (không TM z nguyên dương)

Vay ...