\(\frac{x-15}{100}+\frac{x-10}{105}+\frac{x-5}{110}=\frac{x-100}{15}+\frac{x-105}{10}+\frac{x-110}{5}\)

=> \(\left(\frac{x-15}{100}-1\right)+\left(\frac{x-10}{105}-1\right)+\left(\frac{x-5}{110}-1\right)=\left(\frac{x-100}{15}-1\right)+\)

\(\left(\frac{x-105}{10}-1\right)+\left(\frac{x-110}{5}-1\right)\)

=> \(\frac{x-115}{100}+\frac{x-115}{105}+\frac{x-115}{110}=\frac{x-115}{15}+\frac{x-115}{10}+\frac{x-115}{5}\)

=> \(\left(x-115\right)\left(\frac{1}{100}+\frac{1}{105}+\frac{1}{110}-\frac{1}{15}-\frac{1}{10}-\frac{1}{5}\right)=0\)

=> x - 115 = 0 (Vì \(\frac{1}{100}+\frac{1}{105}+\frac{1}{110}-\frac{1}{15}-\frac{1}{10}-\frac{1}{5}\ne0\)

=> x = 115

Đặt \(V=\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge\frac{a}{b}+\frac{b}{a}\)\(=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\)

Vì a,b dương \(\Rightarrow A\ge\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-2\left(\frac{a}{b}+\frac{b}{a}\right)+2=\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}-1\right)^2\ge0\)

ĐKXĐ : x \(\ne-1\)

Khi đó \(\frac{x-2}{-3x-3}=-4\)

<=> x - 2 = (-4).(-3x - 3)

<=> x - 2 = 12x + 12

<=> 11x = -14

<=> x = -14/11 (tm)

Vậy tập nghiệm phương trình S = \(\left\{-\frac{14}{11}\right\}\)

\(\frac{x-2}{-3x-3}=-4ĐK:x\ne-1\)

\(\Leftrightarrow x-2=-4\left(-3x-3\right)\Leftrightarrow x-2=12x+12\)

\(\Leftrightarrow-11x=14\Leftrightarrow x=-\frac{14}{11}\)

Vậy tập nghiệm phương trình là S = { -14/11 }