(x+3)^2+(3x-8)(3x+8)+1=3[x(x+3)+6] eching

$$( x + 3 ) ^ { 2 } + ( 3 x - 8 ) ( 3 x + 8 ) + 1 = 3 [ x ( x + 3 ) + 6 ]$$

x uchun yechish

$$x=-3$$

$$x = \frac{24}{7} = 3\frac{3}{7} \approx 3,428571429$$

Guruhlash orqali faktorlash qadamlari

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

$$x^{2}+6x+9+\left(3x-8\right)\left(3x+8\right)+1=3\left(x\left(x+3\right)+6\right)$$

Hisoblang: \left(3x-8\right)\left(3x+8\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 8 kvadratini chiqarish.

$$x^{2}+6x+9+\left(3x\right)^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

\left(3x\right)^{2} ni kengaytirish.

$$x^{2}+6x+9+3^{2}x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$x^{2}+6x+9+9x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

10x^{2} ni olish uchun x^{2} va 9x^{2} ni birlashtirish.

$$10x^{2}+6x+9-64+1=3\left(x\left(x+3\right)+6\right)$$

-55 olish uchun 9 dan 64 ni ayirish.

$$10x^{2}+6x-55+1=3\left(x\left(x+3\right)+6\right)$$

-54 olish uchun -55 va 1'ni qo'shing.

$$10x^{2}+6x-54=3\left(x\left(x+3\right)+6\right)$$

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

$$10x^{2}+6x-54=3\left(x^{2}+3x+6\right)$$

3 ga x^{2}+3x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10x^{2}+6x-54=3x^{2}+9x+18$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$10x^{2}+6x-54-3x^{2}=9x+18$$

7x^{2} ni olish uchun 10x^{2} va -3x^{2} ni birlashtirish.

$$7x^{2}+6x-54=9x+18$$

Ikkala tarafdan 9x ni ayirish.

$$7x^{2}+6x-54-9x=18$$

-3x ni olish uchun 6x va -9x ni birlashtirish.

$$7x^{2}-3x-54=18$$

Ikkala tarafdan 18 ni ayirish.

$$7x^{2}-3x-54-18=0$$

-72 olish uchun -54 dan 18 ni ayirish.

$$7x^{2}-3x-72=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 7x^{2}+ax+bx-72 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=7\left(-72\right)=-504$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -504-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-504 2,-252 3,-168 4,-126 6,-84 7,-72 8,-63 9,-56 12,-42 14,-36 18,-28 21,-24$$

Har bir juftlik yigʻindisini hisoblang.

$$1-504=-503 2-252=-250 3-168=-165 4-126=-122 6-84=-78 7-72=-65 8-63=-55 9-56=-47 12-42=-30 14-36=-22 18-28=-10 21-24=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=-24 b=21$$

7x^{2}-3x-72 ni \left(7x^{2}-24x\right)+\left(21x-72\right) sifatida qaytadan yozish.

$$\left(7x^{2}-24x\right)+\left(21x-72\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$x\left(7x-24\right)+3\left(7x-24\right)$$

Distributiv funktsiyasidan foydalangan holda 7x-24 umumiy terminini chiqaring.

$$\left(7x-24\right)\left(x+3\right)$$

Tenglamani yechish uchun 7x-24=0 va x+3=0 ni yeching.

$$x=\frac{24}{7} x=-3$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$x^{2}+6x+9+\left(3x-8\right)\left(3x+8\right)+1=3\left(x\left(x+3\right)+6\right)$$

Hisoblang: \left(3x-8\right)\left(3x+8\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 8 kvadratini chiqarish.

$$x^{2}+6x+9+\left(3x\right)^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

\left(3x\right)^{2} ni kengaytirish.

$$x^{2}+6x+9+3^{2}x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$x^{2}+6x+9+9x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

10x^{2} ni olish uchun x^{2} va 9x^{2} ni birlashtirish.

$$10x^{2}+6x+9-64+1=3\left(x\left(x+3\right)+6\right)$$

-55 olish uchun 9 dan 64 ni ayirish.

$$10x^{2}+6x-55+1=3\left(x\left(x+3\right)+6\right)$$

-54 olish uchun -55 va 1'ni qo'shing.

$$10x^{2}+6x-54=3\left(x\left(x+3\right)+6\right)$$

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

$$10x^{2}+6x-54=3\left(x^{2}+3x+6\right)$$

3 ga x^{2}+3x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10x^{2}+6x-54=3x^{2}+9x+18$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$10x^{2}+6x-54-3x^{2}=9x+18$$

7x^{2} ni olish uchun 10x^{2} va -3x^{2} ni birlashtirish.

$$7x^{2}+6x-54=9x+18$$

Ikkala tarafdan 9x ni ayirish.

$$7x^{2}+6x-54-9x=18$$

-3x ni olish uchun 6x va -9x ni birlashtirish.

$$7x^{2}-3x-54=18$$

Ikkala tarafdan 18 ni ayirish.

$$7x^{2}-3x-54-18=0$$

-72 olish uchun -54 dan 18 ni ayirish.

$$7x^{2}-3x-72=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 7 ni a, -3 ni b va -72 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 7\left(-72\right)}}{2\times 7}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 7\left(-72\right)}}{2\times 7}$$

-4 ni 7 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-28\left(-72\right)}}{2\times 7}$$

-28 ni -72 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+2016}}{2\times 7}$$

9 ni 2016 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{2025}}{2\times 7}$$

2025 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±45}{2\times 7}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±45}{2\times 7}$$

2 ni 7 marotabaga ko'paytirish.

$$x=\frac{3±45}{14}$$

x=\frac{3±45}{14} tenglamasini yeching, bunda ± musbat. 3 ni 45 ga qo'shish.

$$x=\frac{48}{14}$$

\frac{48}{14} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{24}{7}$$

x=\frac{3±45}{14} tenglamasini yeching, bunda ± manfiy. 3 dan 45 ni ayirish.

$$x=-\frac{42}{14}$$

-42 ni 14 ga bo'lish.

$$x=-3$$

Tenglama yechildi.

$$x=\frac{24}{7} x=-3$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$x^{2}+6x+9+\left(3x-8\right)\left(3x+8\right)+1=3\left(x\left(x+3\right)+6\right)$$

Hisoblang: \left(3x-8\right)\left(3x+8\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 8 kvadratini chiqarish.

$$x^{2}+6x+9+\left(3x\right)^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

\left(3x\right)^{2} ni kengaytirish.

$$x^{2}+6x+9+3^{2}x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$x^{2}+6x+9+9x^{2}-64+1=3\left(x\left(x+3\right)+6\right)$$

10x^{2} ni olish uchun x^{2} va 9x^{2} ni birlashtirish.

$$10x^{2}+6x+9-64+1=3\left(x\left(x+3\right)+6\right)$$

-55 olish uchun 9 dan 64 ni ayirish.

$$10x^{2}+6x-55+1=3\left(x\left(x+3\right)+6\right)$$

-54 olish uchun -55 va 1'ni qo'shing.

$$10x^{2}+6x-54=3\left(x\left(x+3\right)+6\right)$$

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

$$10x^{2}+6x-54=3\left(x^{2}+3x+6\right)$$

3 ga x^{2}+3x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10x^{2}+6x-54=3x^{2}+9x+18$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$10x^{2}+6x-54-3x^{2}=9x+18$$

7x^{2} ni olish uchun 10x^{2} va -3x^{2} ni birlashtirish.

$$7x^{2}+6x-54=9x+18$$

Ikkala tarafdan 9x ni ayirish.

$$7x^{2}+6x-54-9x=18$$

-3x ni olish uchun 6x va -9x ni birlashtirish.

$$7x^{2}-3x-54=18$$

54 ni ikki tarafga qo’shing.

$$7x^{2}-3x=18+54$$

72 olish uchun 18 va 54'ni qo'shing.

$$7x^{2}-3x=72$$

Ikki tarafini 7 ga bo‘ling.

$$\frac{7x^{2}-3x}{7}=\frac{72}{7}$$

7 ga bo'lish 7 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{3}{7}x=\frac{72}{7}$$

-\frac{3}{7} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{14} olish uchun. Keyin, -\frac{3}{14} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{3}{7}x+\left(-\frac{3}{14}\right)^{2}=\frac{72}{7}+\left(-\frac{3}{14}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{14} kvadratini chiqarish.

$$x^{2}-\frac{3}{7}x+\frac{9}{196}=\frac{72}{7}+\frac{9}{196}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{72}{7} ni \frac{9}{196} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{3}{7}x+\frac{9}{196}=\frac{2025}{196}$$

x^{2}-\frac{3}{7}x+\frac{9}{196} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{3}{14}\right)^{2}=\frac{2025}{196}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{14}\right)^{2}}=\sqrt{\frac{2025}{196}}$$

Qisqartirish.

$$x-\frac{3}{14}=\frac{45}{14} x-\frac{3}{14}=-\frac{45}{14}$$

\frac{3}{14} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{24}{7} x=-3$$