(x+1)^2-2(x+1)-1leq0 eching

$$( x + 1 ) ^ { 2 } - 2 ( x + 1 ) - 1 \leq 0$$

x uchun yechish

$$x\in \begin{bmatrix}-\sqrt{2},\sqrt{2}\end{bmatrix}$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}+2x+1-2\left(x+1\right)-1\leq 0$$

-2 ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{2}+2x+1-2x-2-1\leq 0$$

0 ni olish uchun 2x va -2x ni birlashtirish.

$$x^{2}+1-2-1\leq 0$$

-1 olish uchun 1 dan 2 ni ayirish.

$$x^{2}-1-1\leq 0$$

-2 olish uchun -1 dan 1 ni ayirish.

$$x^{2}-2\leq 0$$

2 ni ikki tarafga qo’shing.

$$x^{2}\leq 2$$

2 ning kvadrat ildizini hisoblab, \sqrt{2} natijaga ega bo‘ling. 2 ni \left(\sqrt{2}\right)^{2} sifatida qaytadan yozish.

$$x^{2}\leq \left(\sqrt{2}\right)^{2}$$

Tengsizlikda |x|\leq \sqrt{2} bor.

$$|x|\leq \sqrt{2}$$

|x|\leq \sqrt{2} ni x\in \left[-\sqrt{2},\sqrt{2}\right] sifatida qaytadan yozish.

$$x\in \begin{bmatrix}-\sqrt{2},\sqrt{2}\end{bmatrix}$$