3x+2+1/3x+2=7 eching

$$3 x + 2 + \frac { 1 } { 3 x + 2 } = 7$$

x uchun yechish

$$x=\frac{1-\sqrt{5}}{2}\approx -0,618033989$$

$$x = \frac{\sqrt{5} + 1}{2} \approx 1,618033989$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -\frac{2}{3} teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 3x+2 ga ko'paytirish.

$$3x\left(3x+2\right)+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

3x ga 3x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$9x^{2}+6x+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

3x+2 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$9x^{2}+6x+6x+4+1=7\left(3x+2\right)$$

12x ni olish uchun 6x va 6x ni birlashtirish.

$$9x^{2}+12x+4+1=7\left(3x+2\right)$$

5 olish uchun 4 va 1'ni qo'shing.

$$9x^{2}+12x+5=7\left(3x+2\right)$$

7 ga 3x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$9x^{2}+12x+5=21x+14$$

Ikkala tarafdan 21x ni ayirish.

$$9x^{2}+12x+5-21x=14$$

-9x ni olish uchun 12x va -21x ni birlashtirish.

$$9x^{2}-9x+5=14$$

Ikkala tarafdan 14 ni ayirish.

$$9x^{2}-9x+5-14=0$$

-9 olish uchun 5 dan 14 ni ayirish.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, -9 ni b va -9 ni c bilan almashtiring.

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

-9 kvadratini chiqarish.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 9\left(-9\right)}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-36\left(-9\right)}}{2\times 9}$$

-36 ni -9 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81+324}}{2\times 9}$$

81 ni 324 ga qo'shish.

$$x=\frac{-\left(-9\right)±\sqrt{405}}{2\times 9}$$

405 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-9\right)±9\sqrt{5}}{2\times 9}$$

-9 ning teskarisi 9 ga teng.

$$x=\frac{9±9\sqrt{5}}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{9±9\sqrt{5}}{18}$$

x=\frac{9±9\sqrt{5}}{18} tenglamasini yeching, bunda ± musbat. 9 ni 9\sqrt{5} ga qo'shish.

$$x=\frac{9\sqrt{5}+9}{18}$$

9+9\sqrt{5} ni 18 ga bo'lish.

$$x=\frac{\sqrt{5}+1}{2}$$

x=\frac{9±9\sqrt{5}}{18} tenglamasini yeching, bunda ± manfiy. 9 dan 9\sqrt{5} ni ayirish.

$$x=\frac{9-9\sqrt{5}}{18}$$

9-9\sqrt{5} ni 18 ga bo'lish.

$$x=\frac{1-\sqrt{5}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -\frac{2}{3} teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 3x+2 ga ko'paytirish.

$$3x\left(3x+2\right)+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

3x ga 3x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$9x^{2}+6x+\left(3x+2\right)\times 2+1=7\left(3x+2\right)$$

3x+2 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$9x^{2}+6x+6x+4+1=7\left(3x+2\right)$$

12x ni olish uchun 6x va 6x ni birlashtirish.

$$9x^{2}+12x+4+1=7\left(3x+2\right)$$

5 olish uchun 4 va 1'ni qo'shing.

$$9x^{2}+12x+5=7\left(3x+2\right)$$

7 ga 3x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$9x^{2}+12x+5=21x+14$$

Ikkala tarafdan 21x ni ayirish.

$$9x^{2}+12x+5-21x=14$$

-9x ni olish uchun 12x va -21x ni birlashtirish.

$$9x^{2}-9x+5=14$$

Ikkala tarafdan 5 ni ayirish.

$$9x^{2}-9x=14-5$$

9 olish uchun 14 dan 5 ni ayirish.

$$9x^{2}-9x=9$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}-9x}{9}=\frac{9}{9}$$

9 ga bo'lish 9 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{9}{9}\right)x=\frac{9}{9}$$

-9 ni 9 ga bo'lish.

$$x^{2}-x=\frac{9}{9}$$

9 ni 9 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

$$x^{2}-x+\frac{1}{4}=1+\frac{1}{4}$$

1 ni \frac{1}{4} ga qo'shish.

$$x^{2}-x+\frac{1}{4}=\frac{5}{4}$$

x^{2}-x+\frac{1}{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.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Qisqartirish.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$