left(5x+1right)^2-3(5x+1)-4=0 eching

$${ \left(5x+1 \right) }^{ 2 } -3(5x+1)-4=0$$

x uchun yechish

$$x=-\frac{2}{5}=-0,4$$

$$x=\frac{3}{5}=0,6$$

Guruhlash orqali faktorlash qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(5x+1\right)^{2} kengaytirilishi uchun ishlating.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

-3 ga 5x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$25x^{2}+10x+1-15x-3-4=0$$

-5x ni olish uchun 10x va -15x ni birlashtirish.

$$25x^{2}-5x+1-3-4=0$$

-2 olish uchun 1 dan 3 ni ayirish.

$$25x^{2}-5x-2-4=0$$

-6 olish uchun -2 dan 4 ni ayirish.

$$25x^{2}-5x-6=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 25x^{2}+ax+bx-6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=25\left(-6\right)=-150$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -150-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Har bir juftlik yigʻindisini hisoblang.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=-15 b=10$$

25x^{2}-5x-6 ni \left(25x^{2}-15x\right)+\left(10x-6\right) sifatida qaytadan yozish.

$$\left(25x^{2}-15x\right)+\left(10x-6\right)$$

Birinchi guruhda 5x ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$5x\left(5x-3\right)+2\left(5x-3\right)$$

Distributiv funktsiyasidan foydalangan holda 5x-3 umumiy terminini chiqaring.

$$\left(5x-3\right)\left(5x+2\right)$$

Tenglamani yechish uchun 5x-3=0 va 5x+2=0 ni yeching.

$$x=\frac{3}{5} x=-\frac{2}{5}$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(5x+1\right)^{2} kengaytirilishi uchun ishlating.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

-3 ga 5x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$25x^{2}+10x+1-15x-3-4=0$$

-5x ni olish uchun 10x va -15x ni birlashtirish.

$$25x^{2}-5x+1-3-4=0$$

-2 olish uchun 1 dan 3 ni ayirish.

$$25x^{2}-5x-2-4=0$$

-6 olish uchun -2 dan 4 ni ayirish.

$$25x^{2}-5x-6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, -5 ni b va -6 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 25\left(-6\right)}}{2\times 25}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 25\left(-6\right)}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-100\left(-6\right)}}{2\times 25}$$

-100 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+600}}{2\times 25}$$

25 ni 600 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{625}}{2\times 25}$$

625 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±25}{2\times 25}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±25}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$x=\frac{5±25}{50}$$

x=\frac{5±25}{50} tenglamasini yeching, bunda ± musbat. 5 ni 25 ga qo'shish.

$$x=\frac{30}{50}$$

\frac{30}{50} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{3}{5}$$

x=\frac{5±25}{50} tenglamasini yeching, bunda ± manfiy. 5 dan 25 ni ayirish.

$$x=-\frac{20}{50}$$

\frac{-20}{50} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=\frac{3}{5} x=-\frac{2}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(5x+1\right)^{2} kengaytirilishi uchun ishlating.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

-3 ga 5x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$25x^{2}+10x+1-15x-3-4=0$$

-5x ni olish uchun 10x va -15x ni birlashtirish.

$$25x^{2}-5x+1-3-4=0$$

-2 olish uchun 1 dan 3 ni ayirish.

$$25x^{2}-5x-2-4=0$$

-6 olish uchun -2 dan 4 ni ayirish.

$$25x^{2}-5x-6=0$$

6 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$25x^{2}-5x=6$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25x^{2}-5x}{25}=\frac{6}{25}$$

25 ga bo'lish 25 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{5}{25}\right)x=\frac{6}{25}$$

\frac{-5}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{1}{5}x=\frac{6}{25}$$

-\frac{1}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{10} olish uchun. Keyin, -\frac{1}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{6}{25}+\left(-\frac{1}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{10} kvadratini chiqarish.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{6}{25}+\frac{1}{100}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{6}{25} ni \frac{1}{100} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{4}$$

x^{2}-\frac{1}{5}x+\frac{1}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{1}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Qisqartirish.

$$x-\frac{1}{10}=\frac{1}{2} x-\frac{1}{10}=-\frac{1}{2}$$

\frac{1}{10} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3}{5} x=-\frac{2}{5}$$