(2x-1)^2geq(2x+3)^2 eching

$$( 2 x - 1 ) ^ { 2 } \geq ( 2 x + 3 ) ^ { 2 }$$

x uchun yechish

$$x\leq -\frac{1}{2}$$

Yechim qadamlari

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

$$4x^{2}-4x+1\geq \left(2x+3\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2x+3\right)^{2} kengaytirilishi uchun ishlating.

$$4x^{2}-4x+1\geq 4x^{2}+12x+9$$

Ikkala tarafdan 4x^{2} ni ayirish.

$$4x^{2}-4x+1-4x^{2}\geq 12x+9$$

0 ni olish uchun 4x^{2} va -4x^{2} ni birlashtirish.

$$-4x+1\geq 12x+9$$

Ikkala tarafdan 12x ni ayirish.

$$-4x+1-12x\geq 9$$

-16x ni olish uchun -4x va -12x ni birlashtirish.

$$-16x+1\geq 9$$

Ikkala tarafdan 1 ni ayirish.

$$-16x\geq 9-1$$

8 olish uchun 9 dan 1 ni ayirish.

$$-16x\geq 8$$

Ikki tarafini -16 ga bo‘ling. -16 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$x\leq \frac{8}{-16}$$

\frac{8}{-16} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x\leq -\frac{1}{2}$$