4x+6/12x+4=2x/6 eching

$$\frac { 4 x + 6 } { 12 x + 4 } = \frac { 2 x } { 6 }$$

x uchun yechish

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3}\approx 1,602628851$$

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3}\approx -0,935962184$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -\frac{1}{3} teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 12\left(3x+1\right) ga, 12x+4,6 ning eng kichik karralisiga ko‘paytiring.

$$3\left(4x+6\right)=\left(6x+2\right)\times 2x$$

3 ga 4x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x+18=\left(6x+2\right)\times 2x$$

6x+2 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x+18=\left(12x+4\right)x$$

12x+4 ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x+18=12x^{2}+4x$$

Ikkala tarafdan 12x^{2} ni ayirish.

$$12x+18-12x^{2}=4x$$

Ikkala tarafdan 4x ni ayirish.

$$12x+18-12x^{2}-4x=0$$

8x ni olish uchun 12x va -4x ni birlashtirish.

$$8x+18-12x^{2}=0$$

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.

$$-12x^{2}+8x+18=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -12 ni a, 8 ni b va 18 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-12\right)\times 18}}{2\left(-12\right)}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\left(-12\right)\times 18}}{2\left(-12\right)}$$

-4 ni -12 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+48\times 18}}{2\left(-12\right)}$$

48 ni 18 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+864}}{2\left(-12\right)}$$

64 ni 864 ga qo'shish.

$$x=\frac{-8±\sqrt{928}}{2\left(-12\right)}$$

928 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±4\sqrt{58}}{2\left(-12\right)}$$

2 ni -12 marotabaga ko'paytirish.

$$x=\frac{-8±4\sqrt{58}}{-24}$$

x=\frac{-8±4\sqrt{58}}{-24} tenglamasini yeching, bunda ± musbat. -8 ni 4\sqrt{58} ga qo'shish.

$$x=\frac{4\sqrt{58}-8}{-24}$$

-8+4\sqrt{58} ni -24 ga bo'lish.

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3}$$

x=\frac{-8±4\sqrt{58}}{-24} tenglamasini yeching, bunda ± manfiy. -8 dan 4\sqrt{58} ni ayirish.

$$x=\frac{-4\sqrt{58}-8}{-24}$$

-8-4\sqrt{58} ni -24 ga bo'lish.

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Tenglama yechildi.

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3} x=\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -\frac{1}{3} teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 12\left(3x+1\right) ga, 12x+4,6 ning eng kichik karralisiga ko‘paytiring.

$$3\left(4x+6\right)=\left(6x+2\right)\times 2x$$

3 ga 4x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x+18=\left(6x+2\right)\times 2x$$

6x+2 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x+18=\left(12x+4\right)x$$

12x+4 ga x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x+18=12x^{2}+4x$$

Ikkala tarafdan 12x^{2} ni ayirish.

$$12x+18-12x^{2}=4x$$

Ikkala tarafdan 4x ni ayirish.

$$12x+18-12x^{2}-4x=0$$

8x ni olish uchun 12x va -4x ni birlashtirish.

$$8x+18-12x^{2}=0$$

Ikkala tarafdan 18 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$8x-12x^{2}=-18$$

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

$$-12x^{2}+8x=-18$$

Ikki tarafini -12 ga bo‘ling.

$$\frac{-12x^{2}+8x}{-12}=-\frac{18}{-12}$$

-12 ga bo'lish -12 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{8}{-12}x=-\frac{18}{-12}$$

\frac{8}{-12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{2}{3}x=-\frac{18}{-12}$$

\frac{-18}{-12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{3} kvadratini chiqarish.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{3}{2}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{2} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{29}{18}$$

x^{2}-\frac{2}{3}x+\frac{1}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{29}{18}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{29}{18}}$$

Qisqartirish.

$$x-\frac{1}{3}=\frac{\sqrt{58}}{6} x-\frac{1}{3}=-\frac{\sqrt{58}}{6}$$

\frac{1}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3} x=-\frac{\sqrt{58}}{6}+\frac{1}{3}$$