x(x+3)/2-(x+1)^2/3+1/3=0 eching

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

x uchun yechish

$$x=-5$$

$$x=0$$

Faktorlash qadamlari

Tenglamaning ikkala tarafini 6 ga, 2,3 ning eng kichik karralisiga ko‘paytiring.

$$3x\left(x+3\right)-2\left(x+1\right)^{2}+2=0$$

3x ga x+3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}+9x-2\left(x+1\right)^{2}+2=0$$

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

$$3x^{2}+9x-2\left(x^{2}+2x+1\right)+2=0$$

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

$$3x^{2}+9x-2x^{2}-4x-2+2=0$$

x^{2} ni olish uchun 3x^{2} va -2x^{2} ni birlashtirish.

$$x^{2}+9x-4x-2+2=0$$

5x ni olish uchun 9x va -4x ni birlashtirish.

$$x^{2}+5x-2+2=0$$

0 olish uchun -2 va 2'ni qo'shing.

$$x^{2}+5x=0$$

x omili.

$$x\left(x+5\right)=0$$

Tenglamani yechish uchun x=0 va x+5=0 ni yeching.

$$x=0 x=-5$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tenglamaning ikkala tarafini 6 ga, 2,3 ning eng kichik karralisiga ko‘paytiring.

$$3x\left(x+3\right)-2\left(x+1\right)^{2}+2=0$$

3x ga x+3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}+9x-2\left(x+1\right)^{2}+2=0$$

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

$$3x^{2}+9x-2\left(x^{2}+2x+1\right)+2=0$$

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

$$3x^{2}+9x-2x^{2}-4x-2+2=0$$

x^{2} ni olish uchun 3x^{2} va -2x^{2} ni birlashtirish.

$$x^{2}+9x-4x-2+2=0$$

5x ni olish uchun 9x va -4x ni birlashtirish.

$$x^{2}+5x-2+2=0$$

0 olish uchun -2 va 2'ni qo'shing.

$$x^{2}+5x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 5 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}}}{2}$$

5^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-5±5}{2}$$

x=\frac{-5±5}{2} tenglamasini yeching, bunda ± musbat. -5 ni 5 ga qo'shish.

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

0 ni 2 ga bo'lish.

$$x=0$$

x=\frac{-5±5}{2} tenglamasini yeching, bunda ± manfiy. -5 dan 5 ni ayirish.

$$x=-\frac{10}{2}$$

-10 ni 2 ga bo'lish.

$$x=-5$$

Tenglama yechildi.

$$x=0 x=-5$$

Kvadratni yakunlash bo'yicha qadamlar

Tenglamaning ikkala tarafini 6 ga, 2,3 ning eng kichik karralisiga ko‘paytiring.

$$3x\left(x+3\right)-2\left(x+1\right)^{2}+2=0$$

3x ga x+3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}+9x-2\left(x+1\right)^{2}+2=0$$

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

$$3x^{2}+9x-2\left(x^{2}+2x+1\right)+2=0$$

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

$$3x^{2}+9x-2x^{2}-4x-2+2=0$$

x^{2} ni olish uchun 3x^{2} va -2x^{2} ni birlashtirish.

$$x^{2}+9x-4x-2+2=0$$

5x ni olish uchun 9x va -4x ni birlashtirish.

$$x^{2}+5x-2+2=0$$

0 olish uchun -2 va 2'ni qo'shing.

$$x^{2}+5x=0$$

5 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{2} olish uchun. Keyin, \frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=\left(\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

$$x^{2}+5x+\frac{25}{4}=\frac{25}{4}$$

x^{2}+5x+\frac{25}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Qisqartirish.

$$x+\frac{5}{2}=\frac{5}{2} x+\frac{5}{2}=-\frac{5}{2}$$

Tenglamaning ikkala tarafidan \frac{5}{2} ni ayirish.

$$x=0 x=-5$$