Lutasin ang x^2geq12^21

$$x ^ { 2 } \geq 12 ^ { 21 }$$

I-solve ang x

$$x\in (-\infty,-123834728448\sqrt{3}]\cup [123834728448\sqrt{3},\infty)$$

Mga Hakbang ng Solution

Kalkulahin ang 12 sa power ng 21 at kunin ang 46005119909369701466112.

$$x^{2}\geq 46005119909369701466112$$

Kalkulahin ang square root ng 46005119909369701466112 at makuha ang 123834728448\sqrt{3}. I-rewrite ang 46005119909369701466112 bilang \left(123834728448\sqrt{3}\right)^{2}.

$$x^{2}\geq \left(123834728448\sqrt{3}\right)^{2}$$

Nalalapat ang inequality para sa |x|\geq 123834728448\sqrt{3}.

$$|x|\geq 123834728448\sqrt{3}$$

I-rewrite ang |x|\geq 123834728448\sqrt{3} bilang x\leq -123834728448\sqrt{3}\text{; }x\geq 123834728448\sqrt{3}.

$$x\leq -123834728448\sqrt{3}\text{; }x\geq 123834728448\sqrt{3}$$