(5x-1)^2=16 eching

$$( 5 x - 1 ) ^ { 2 } = 16$$

x uchun yechish

$$x=1$$

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

Guruhlash orqali faktorlash qadamlari

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

$$25x^{2}-10x+1=16$$

Ikkala tarafdan 16 ni ayirish.

$$25x^{2}-10x+1-16=0$$

-15 olish uchun 1 dan 16 ni ayirish.

$$25x^{2}-10x-15=0$$

Ikki tarafini 5 ga bo‘ling.

$$5x^{2}-2x-3=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5x^{2}+ax+bx-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-2 ab=5\left(-3\right)=-15$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -15-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-15 3,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-15=-14 3-5=-2$$

Yechim – -2 yigʻindisini beruvchi juftlik.

$$a=-5 b=3$$

5x^{2}-2x-3 ni \left(5x^{2}-5x\right)+\left(3x-3\right) sifatida qaytadan yozish.

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

Birinchi guruhda 5x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$5x\left(x-1\right)+3\left(x-1\right)$$

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

$$\left(x-1\right)\left(5x+3\right)$$

Tenglamani yechish uchun x-1=0 va 5x+3=0 ni yeching.

$$x=1 x=-\frac{3}{5}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$25x^{2}-10x+1=16$$

Ikkala tarafdan 16 ni ayirish.

$$25x^{2}-10x+1-16=0$$

-15 olish uchun 1 dan 16 ni ayirish.

$$25x^{2}-10x-15=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, -10 ni b va -15 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25\left(-15\right)}}{2\times 25}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25\left(-15\right)}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-100\left(-15\right)}}{2\times 25}$$

-100 ni -15 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+1500}}{2\times 25}$$

100 ni 1500 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{1600}}{2\times 25}$$

1600 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±40}{2\times 25}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±40}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$x=\frac{10±40}{50}$$

x=\frac{10±40}{50} tenglamasini yeching, bunda ± musbat. 10 ni 40 ga qo'shish.

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

50 ni 50 ga bo'lish.

$$x=1$$

x=\frac{10±40}{50} tenglamasini yeching, bunda ± manfiy. 10 dan 40 ni ayirish.

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

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

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

Tenglama yechildi.

$$x=1 x=-\frac{3}{5}$$

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.

$$25x^{2}-10x+1=16$$

Ikkala tarafdan 1 ni ayirish.

$$25x^{2}-10x=16-1$$

15 olish uchun 16 dan 1 ni ayirish.

$$25x^{2}-10x=15$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25x^{2}-10x}{25}=\frac{15}{25}$$

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

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

\frac{-10}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{2}{5}x=\frac{15}{25}$$

\frac{15}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{5} kvadratini chiqarish.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{3}{5}+\frac{1}{25}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{5} ni \frac{1}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{16}{25}$$

x^{2}-\frac{2}{5}x+\frac{1}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{16}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Qisqartirish.

$$x-\frac{1}{5}=\frac{4}{5} x-\frac{1}{5}=-\frac{4}{5}$$

\frac{1}{5} ni tenglamaning ikkala tarafiga qo'shish.

$$x=1 x=-\frac{3}{5}$$