-3x(2-x+4x)=1 eching

$$- 3 x ( 2 - x + 4 x ) = 1$$

x uchun yechish

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

Kvadrat tenglamalaridan foydalanish qoidalari

3x ni olish uchun -x va 4x ni birlashtirish.

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

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

$$-6x-9x^{2}=1$$

Ikkala tarafdan 1 ni ayirish.

$$-6x-9x^{2}-1=0$$

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.

$$-9x^{2}-6x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -9 ni a, -6 ni b va -1 ni c bilan almashtiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

-4 ni -9 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+36\left(-1\right)}}{2\left(-9\right)}$$

36 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2\left(-9\right)}$$

36 ni -36 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\left(-9\right)}$$

0 ning kvadrat ildizini chiqarish.

$$x=-\frac{-6}{2\left(-9\right)}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6}{2\left(-9\right)}$$

2 ni -9 marotabaga ko'paytirish.

$$x=\frac{6}{-18}$$

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

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

Kvadratni yakunlash bo'yicha qadamlar

3x ni olish uchun -x va 4x ni birlashtirish.

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

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

$$-6x-9x^{2}=1$$

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

$$-9x^{2}-6x=1$$

Ikki tarafini -9 ga bo‘ling.

$$\frac{-9x^{2}-6x}{-9}=\frac{1}{-9}$$

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

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

\frac{-6}{-9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

1 ni -9 ga bo'lish.

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

\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.

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

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

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

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=0$$

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{1}{3} ni ayirish.

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

Tenglama yechildi. Yechimlar bir xil.

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