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bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
\(x^8+x^8+y^8+y^8+y^8+z^8+z^8+z^8\ge8\sqrt[8]{x^{16}y^{24}z^{24}}=8x^2y^3z^3\)
Tương tự: \(3x^8+2y^8+3z^8\ge8x^3y^2z^3\)
\(3x^8+3y^8+2z^8\ge8x^3y^3z^2\)
Cộng vế với vế:
\(8\left(x^8+y^8+z^8\right)\ge8\left(x^2y^3z^3+x^3y^2z^3+x^3y^3z^2\right)\)
\(\Leftrightarrow\frac{x^8+y^8+z^8}{x^3y^3z^3}\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Dấu "=" xảy ra khi \(x=y=z\)
\(H=\frac{a^2}{2017a^2+a}+\frac{b^2}{2017b^2+b}+\frac{c^2}{2017c^2+c}\ge\frac{\left(a+b+c\right)^2}{2017\left(a^2+b^2+c^2\right)+\left(a+b+c\right)}\)
\(H\ge\frac{\left(a+b+c\right)^2}{2017.\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)}=\frac{1}{\frac{2017}{3}+1}=\frac{3}{2020}\)
\(\Rightarrow H_{max}=\frac{3}{2020}\) khi \(a=b=c=\frac{1}{3}\)
