$$ \left(\frac{2}{7}\right)^{\frac{2 x+1}{x+2}} \geq 3,5 $$ tengsizlikni yechimi bo'ladigan oraliq uzunligini toping.

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Chuqur tahlil qilamiz. Birinchi qadamda, tengsizlikni quyidagi shaklga yozamiz: $$ \left(\frac{2}{7}\right)^{\frac{2x+1}{x+2}} \ge \,3,5. $$ E'tibor bering, $$3,5$$ ni kasr shakliga o‘tkazsak: $$ 3,5=\frac{7}{2}. $$ Shunday qilib, tengsizlik $$ \left(\frac{2}{7}\right)^{\frac{2x+1}{x+2}} \ge \frac{7}{2} $$ holatga keladi. Endi, quyidagi kuzatmani amalga oshiramiz: Bizda asos $$ \frac{2}{7} $$ $$ (0<\frac{2}{7}<1) $$ ga teng. Shu sababli, $$ \left(\frac{2}{7}\right)^t $$ funksiya kamayuvchi bo‘ladi. Bundan kelib chiqadiki, agar $$ \left(\frac{2}{7}\right)^{\alpha} \ge \left(\frac{2}{7}\right)^{\beta}, $$ unda aslida $$ \alpha\le \beta. $$ Shuningdek, e'tibor bering: $$ \left(\frac{2}{7}\right)^{-1}=\frac{7}{2}. $$ Demak tengsizlikni $$ \left(\frac{2}{7}\right)^{\frac{2x+1}{x+2}} \ge \left(\frac{2}{7}\right)^{-1} $$ deb yozsak, eksponentlarga nisbatan quyidagi tengsizlik hosil bo‘ladi: $$ \frac{2x+1}{x+2} \le -1. $$ Biz ushbu tengsizlikni yechamiz. Birinchi bosqichda, tengsizlikning ikkala tomoniga $$1$$ qo‘shib olamiz: $$ \frac{2x+1}{x+2} + 1 \le 0. $$ Qayta yozamiz: $$ \frac{2x+1 + (x+2)}{x+2} \le 0 \quad \Longrightarrow \quad \frac{3x+3}{x+2} \le 0. $$ Chuqurlashtirib, numeratorni omillar ko‘rinishiga keltiramiz: $$ \frac{3(x+1)}{x+2} \le 0. $$ Musbat $$3$$ ga bolish orqali tengsizlik o‘zgarmaydi: $$ \frac{x+1}{x+2} \le 0. $$ Endi, bu kasrli ifodaning belgisini aniqlash uchun kritik nuqtalarni topamiz: - $$x+1=0 \quad \Rightarrow \quad x=-1,$$ - $$x+2=0 \quad \Rightarrow \quad x=-2.$$ Shuningdek, $$x=-2$$ nuqtasida ifoda aniqlanmaganligini sezamiz (bo‘linma nolga teng bo‘ladi). Intervallarni aniqlaymiz: - Agar $$x<-2$$, masalan $$x=-3$$: $$ (-3)+1=-2\ (<0), \quad (-3)+2=-1\ (<0), \quad \text{shunday qilib, } \frac{-2}{-1}=2>0. $$ - Agar $$-20), \quad \text{shunday qilib, } \frac{-0.5}{0.5}=-1<0. \] - Agar $$x>-1$$, masalan $$x=0$$: $$ 0+1=1\ (>0), \quad 0+2=2\ (>0), \quad \text{shunday qilib, } \frac{1}{2}=0.5>0. $$ Shu tahlil natijasida, $$\frac{x+1}{x+2}$$ ifodasi faqat $$-2Tenglama yechim oraliq uzunligi 1 ga teng.

tengsizlikni yechimi bo’ladigan
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tengsizlikni yechimi bo’ladigan oraliq uzunligini toping.