3(1-x)+4(1+2x)(1-x)=7 eching

$$3(1-x)+4(1+2x)(1-x)=7$$

x uchun yechish

$$x=\frac{1}{8}=0,125$$

$$x=0$$

Kvadrat tenglamalaridan foydalanish qoidalari

3 ga 1-x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3-3x+4\left(1+2x\right)\left(1-x\right)=7$$

4 ga 1+2x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3-3x+\left(4+8x\right)\left(1-x\right)=7$$

4+8x ga 1-x ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$3-3x+4+4x-8x^{2}=7$$

7 olish uchun 3 va 4'ni qo'shing.

$$7-3x+4x-8x^{2}=7$$

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

$$7+x-8x^{2}=7$$

Ikkala tarafdan 7 ni ayirish.

$$7+x-8x^{2}-7=0$$

0 olish uchun 7 dan 7 ni ayirish.

$$x-8x^{2}=0$$

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -8 ni a, 1 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-8\right)}$$

1^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-1±1}{2\left(-8\right)}$$

2 ni -8 marotabaga ko'paytirish.

$$x=\frac{-1±1}{-16}$$

x=\frac{-1±1}{-16} tenglamasini yeching, bunda ± musbat. -1 ni 1 ga qo'shish.

$$x=\frac{0}{-16}$$

0 ni -16 ga bo'lish.

$$x=0$$

x=\frac{-1±1}{-16} tenglamasini yeching, bunda ± manfiy. -1 dan 1 ni ayirish.

$$x=-\frac{2}{-16}$$

\frac{-2}{-16} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{8}$$

Tenglama yechildi.

$$x=0 x=\frac{1}{8}$$

Kvadratni yakunlash bo'yicha qadamlar

3 ga 1-x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3-3x+4\left(1+2x\right)\left(1-x\right)=7$$

4 ga 1+2x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3-3x+\left(4+8x\right)\left(1-x\right)=7$$

4+8x ga 1-x ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$3-3x+4+4x-8x^{2}=7$$

7 olish uchun 3 va 4'ni qo'shing.

$$7-3x+4x-8x^{2}=7$$

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

$$7+x-8x^{2}=7$$

Ikkala tarafdan 7 ni ayirish.

$$x-8x^{2}=7-7$$

0 olish uchun 7 dan 7 ni ayirish.

$$x-8x^{2}=0$$

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

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

Ikki tarafini -8 ga bo‘ling.

$$\frac{-8x^{2}+x}{-8}=\frac{0}{-8}$$

-8 ga bo'lish -8 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{1}{-8}x=\frac{0}{-8}$$

1 ni -8 ga bo'lish.

$$x^{2}-\frac{1}{8}x=\frac{0}{-8}$$

0 ni -8 ga bo'lish.

$$x^{2}-\frac{1}{8}x=0$$

-\frac{1}{8} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{16} olish uchun. Keyin, -\frac{1}{16} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\left(-\frac{1}{16}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{16} kvadratini chiqarish.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{1}{256}$$

x^{2}-\frac{1}{8}x+\frac{1}{256} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{1}{256}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Qisqartirish.

$$x-\frac{1}{16}=\frac{1}{16} x-\frac{1}{16}=-\frac{1}{16}$$

\frac{1}{16} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1}{8} x=0$$