1) f(x)=0\Rightarrow $5^2+2x+1=0$\Rightarrow $2x+26=0$\Rightarrow $x=-13$
2) Ta có: $h(x)=x^2+x+1=x^2+\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{4}+\dfrac{3}{4}$
\Rightarrow $h(x)=x(x+\dfrac{1}{2})+\dfrac{1}{2}(x+\dfrac{1}{2})+\dfrac{3}{4}$
\Rightarrow $h(x)=(x+\dfrac{1}{2})^2+\dfrac{3}{4}$
Vì $(x+\dfrac{1}{2})^2$\geq 0.\Rightarrow h(x)\geq $\dfrac{3}{4}$
Vậy GTNN của h(x) là $\dfrac{3}{4}$ khi $(x+\dfrac{1}{2})^2=0$\Leftrightarrow $x=-\frac{1}{2}$.
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