\(x+2x+3x+...+10x=550\)

\(\Leftrightarrow (1+2+3+...+10)x=550\)

\(\Leftrightarrow 55x=550\)

\(\Leftrightarrow x=10\)

Đáp án \(A\)

\(x^3+4x=0\)

\(\Leftrightarrow x(x^2+4)=0\)

\(\Leftrightarrow \left[\begin{array}{l}x=0\\x^2+4=0\end{array}\right.\)

\(\Leftrightarrow \left[\begin{array}{l}x=0\ (tm)\\ x^2=-4\ (ktm)\end{array}\right.\)

Vậy \(S=\{0\}\)

\(\to\) Đa thức chỉ có \(1\) nghiệm

\((a+b)\left(\dfrac1a+\dfrac1b\right)\ge 4\)

\(\Leftrightarrow \dfrac1a+\dfrac1b \ge \dfrac{4}{a+b}\)

\(\Leftrightarrow \dfrac{(a+b)b}{(a+b)ab}+\dfrac{(a+b)a}{(a+b)ab}\ge \dfrac{4ab}{(a+b)ab}\)

\(\Leftrightarrow (a+b)b+(a+b)a \ge 4ab\)

\(\Leftrightarrow (a+b)^2 \ge 4ab\)

\(\Leftrightarrow (a+b)^2 - 4ab \ge 0\)

\(\Leftrightarrow a^2+2ab+b^2 -4ab \ge 0\)

\(\Leftrightarrow a^2-2ab+b^2 \ge 0\)

\(\Leftrightarrow (a-b)^2 \ge 0\)

Vậy bất đẳng thức đã cho luôn đúng 

Dấu \("="\) xảy ra khi \(a=b\)