x+1=3x^2+1+0 eching

$$x + 1 = 3 x ^ { 2 } + 1 + 0$$

x uchun yechish

$$x=\frac{1}{3}\approx 0,333333333$$

$$x=0$$

Faktorlash qadamlari

1 olish uchun 1 va 0'ni qo'shing.

$$x+1=3x^{2}+1$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$x+1-3x^{2}=1$$

Ikkala tarafdan 1 ni ayirish.

$$x+1-3x^{2}-1=0$$

0 olish uchun 1 dan 1 ni ayirish.

$$x-3x^{2}=0$$

x omili.

$$x\left(1-3x\right)=0$$

Tenglamani yechish uchun x=0 va 1-3x=0 ni yeching.

$$x=0 x=\frac{1}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

1 olish uchun 1 va 0'ni qo'shing.

$$x+1=3x^{2}+1$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$x+1-3x^{2}=1$$

Ikkala tarafdan 1 ni ayirish.

$$x+1-3x^{2}-1=0$$

0 olish uchun 1 dan 1 ni ayirish.

$$x-3x^{2}=0$$

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.

$$-3x^{2}+x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -3 ni a, 1 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-3\right)}$$

1^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-1±1}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{-1±1}{-6}$$

x=\frac{-1±1}{-6} tenglamasini yeching, bunda ± musbat. -1 ni 1 ga qo'shish.

$$x=\frac{0}{-6}$$

0 ni -6 ga bo'lish.

$$x=0$$

x=\frac{-1±1}{-6} tenglamasini yeching, bunda ± manfiy. -1 dan 1 ni ayirish.

$$x=-\frac{2}{-6}$$

\frac{-2}{-6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{3}$$

Tenglama yechildi.

$$x=0 x=\frac{1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

1 olish uchun 1 va 0'ni qo'shing.

$$x+1=3x^{2}+1$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$x+1-3x^{2}=1$$

Ikkala tarafdan 1 ni ayirish.

$$x-3x^{2}=1-1$$

0 olish uchun 1 dan 1 ni ayirish.

$$x-3x^{2}=0$$

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

$$-3x^{2}+x=0$$

Ikki tarafini -3 ga bo‘ling.

$$\frac{-3x^{2}+x}{-3}=\frac{0}{-3}$$

-3 ga bo'lish -3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{1}{-3}x=\frac{0}{-3}$$

1 ni -3 ga bo'lish.

$$x^{2}-\frac{1}{3}x=\frac{0}{-3}$$

0 ni -3 ga bo'lish.

$$x^{2}-\frac{1}{3}x=0$$

-\frac{1}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{6} olish uchun. Keyin, -\frac{1}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\left(-\frac{1}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{6} kvadratini chiqarish.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

x^{2}-\frac{1}{3}x+\frac{1}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Qisqartirish.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

\frac{1}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1}{3} x=0$$