(2x+1)^2=1+(x-1)(x+1) eching

$$( 2 x + 1 ) ^ { 2 } = 1 + ( x - 1 ) ( x + 1 )$$

x uchun yechish

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

$$x=-1$$

Guruhlash orqali faktorlash qadamlari

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

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$4x^{2}+4x+1=1+x^{2}-1$$

0 olish uchun 1 dan 1 ni ayirish.

$$4x^{2}+4x+1=x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

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

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

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

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx+1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=4 ab=3\times 1=3$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. Faqat bundan juftlik tizim yechimidir.

$$a=1 b=3$$

3x^{2}+4x+1 ni \left(3x^{2}+x\right)+\left(3x+1\right) sifatida qaytadan yozish.

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

3x^{2}+x ichida x ni ajrating.

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

Distributiv funktsiyasidan foydalangan holda 3x+1 umumiy terminini chiqaring.

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

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

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

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$4x^{2}+4x+1=1+x^{2}-1$$

0 olish uchun 1 dan 1 ni ayirish.

$$4x^{2}+4x+1=x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

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

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 4 ni b va 1 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2\times 3}$$

4 kvadratini chiqarish.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-4±\sqrt{16-12}}{2\times 3}$$

16 ni -12 ga qo'shish.

$$x=\frac{-4±\sqrt{4}}{2\times 3}$$

4 ning kvadrat ildizini chiqarish.

$$x=\frac{-4±2}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-4±2}{6}$$

x=\frac{-4±2}{6} tenglamasini yeching, bunda ± musbat. -4 ni 2 ga qo'shish.

$$x=-\frac{2}{6}$$

\frac{-2}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{-4±2}{6} tenglamasini yeching, bunda ± manfiy. -4 dan 2 ni ayirish.

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

-6 ni 6 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

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

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$4x^{2}+4x+1=1+x^{2}-1$$

0 olish uchun 1 dan 1 ni ayirish.

$$4x^{2}+4x+1=x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

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

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

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

Ikkala tarafdan 1 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+4x}{3}=-\frac{1}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{4}{3}x=-\frac{1}{3}$$

\frac{4}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{3} olish uchun. Keyin, \frac{2}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{3} kvadratini chiqarish.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{3} ni \frac{4}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

x^{2}+\frac{4}{3}x+\frac{4}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{2}{3} ni ayirish.

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