x(1+6x)-2x(1+2x)=0 eching

$$x(1+6x)-2x(1+2x)=0$$

x uchun yechish

$$x=\frac{1}{2}=0,5$$

$$x=0$$

Faktorlash qadamlari

x ga 1+6x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+6x^{2}-2x\left(1+2x\right)=0$$

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

$$x+6x^{2}-2x-4x^{2}=0$$

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

$$-x+6x^{2}-4x^{2}=0$$

2x^{2} ni olish uchun 6x^{2} va -4x^{2} ni birlashtirish.

$$-x+2x^{2}=0$$

x omili.

$$x\left(-1+2x\right)=0$$

Tenglamani yechish uchun x=0 va -1+2x=0 ni yeching.

$$x=0 x=\frac{1}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

x ga 1+6x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+6x^{2}-2x\left(1+2x\right)=0$$

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

$$x+6x^{2}-2x-4x^{2}=0$$

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

$$-x+6x^{2}-4x^{2}=0$$

2x^{2} ni olish uchun 6x^{2} va -4x^{2} ni birlashtirish.

$$-x+2x^{2}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$2x^{2}-x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -1 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\times 2}$$

1 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±1}{2\times 2}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±1}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{1±1}{4}$$

x=\frac{1±1}{4} tenglamasini yeching, bunda ± musbat. 1 ni 1 ga qo'shish.

$$x=\frac{2}{4}$$

\frac{2}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{2}$$

x=\frac{1±1}{4} tenglamasini yeching, bunda ± manfiy. 1 dan 1 ni ayirish.

$$x=\frac{0}{4}$$

0 ni 4 ga bo'lish.

$$x=0$$

Tenglama yechildi.

$$x=\frac{1}{2} x=0$$

Kvadratni yakunlash bo'yicha qadamlar

x ga 1+6x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x+6x^{2}-2x\left(1+2x\right)=0$$

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

$$x+6x^{2}-2x-4x^{2}=0$$

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

$$-x+6x^{2}-4x^{2}=0$$

2x^{2} ni olish uchun 6x^{2} va -4x^{2} ni birlashtirish.

$$-x+2x^{2}=0$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$2x^{2}-x=0$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-x}{2}=\frac{0}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{1}{2}x=\frac{0}{2}$$

0 ni 2 ga bo'lish.

$$x^{2}-\frac{1}{2}x=0$$

-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

x^{2}-\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Qisqartirish.

$$x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}$$

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1}{2} x=0$$