(3x-1)(3x+1)-(5x-4)(2x+3)=3 eching

$$( 3 x - 1 ) ( 3 x + 1 ) - ( 5 x - 4 ) ( 2 x + 3 ) = 3$$

x uchun yechish

$$x=-8$$

$$x=1$$

Kvadrat tenglamalaridan foydalanish qoidalari

Hisoblang: \left(3x-1\right)\left(3x+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.

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

\left(3x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$9x^{2}-1-\left(5x-4\right)\left(2x+3\right)=3$$

5x-4 ga 2x+3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$9x^{2}-1-\left(10x^{2}+7x-12\right)=3$$

10x^{2}+7x-12 teskarisini topish uchun har birining teskarisini toping.

$$9x^{2}-1-10x^{2}-7x+12=3$$

-x^{2} ni olish uchun 9x^{2} va -10x^{2} ni birlashtirish.

$$-x^{2}-1-7x+12=3$$

11 olish uchun -1 va 12'ni qo'shing.

$$-x^{2}+11-7x=3$$

Ikkala tarafdan 3 ni ayirish.

$$-x^{2}+11-7x-3=0$$

8 olish uchun 11 dan 3 ni ayirish.

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

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -7 ni b va 8 ni c bilan almashtiring.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

-7 kvadratini chiqarish.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\times 8}}{2\left(-1\right)}$$

4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\left(-1\right)}$$

49 ni 32 ga qo'shish.

$$x=\frac{-\left(-7\right)±\sqrt{81}}{2\left(-1\right)}$$

81 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-7\right)±9}{2\left(-1\right)}$$

-7 ning teskarisi 7 ga teng.

$$x=\frac{7±9}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{7±9}{-2}$$

x=\frac{7±9}{-2} tenglamasini yeching, bunda ± musbat. 7 ni 9 ga qo'shish.

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

16 ni -2 ga bo'lish.

$$x=-8$$

x=\frac{7±9}{-2} tenglamasini yeching, bunda ± manfiy. 7 dan 9 ni ayirish.

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

-2 ni -2 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=-8 x=1$$

Kvadratni yakunlash bo'yicha qadamlar

Hisoblang: \left(3x-1\right)\left(3x+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.

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

\left(3x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 3 ga hisoblang va 9 ni qiymatni oling.

$$9x^{2}-1-\left(5x-4\right)\left(2x+3\right)=3$$

5x-4 ga 2x+3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$9x^{2}-1-\left(10x^{2}+7x-12\right)=3$$

10x^{2}+7x-12 teskarisini topish uchun har birining teskarisini toping.

$$9x^{2}-1-10x^{2}-7x+12=3$$

-x^{2} ni olish uchun 9x^{2} va -10x^{2} ni birlashtirish.

$$-x^{2}-1-7x+12=3$$

11 olish uchun -1 va 12'ni qo'shing.

$$-x^{2}+11-7x=3$$

Ikkala tarafdan 11 ni ayirish.

$$-x^{2}-7x=3-11$$

-8 olish uchun 3 dan 11 ni ayirish.

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

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}-7x}{-1}=-\frac{8}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{7}{-1}\right)x=-\frac{8}{-1}$$

-7 ni -1 ga bo'lish.

$$x^{2}+7x=-\frac{8}{-1}$$

-8 ni -1 ga bo'lish.

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

7 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{7}{2} olish uchun. Keyin, \frac{7}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=8+\left(\frac{7}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{2} kvadratini chiqarish.

$$x^{2}+7x+\frac{49}{4}=8+\frac{49}{4}$$

8 ni \frac{49}{4} ga qo'shish.

$$x^{2}+7x+\frac{49}{4}=\frac{81}{4}$$

x^{2}+7x+\frac{49}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{81}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Qisqartirish.

$$x+\frac{7}{2}=\frac{9}{2} x+\frac{7}{2}=-\frac{9}{2}$$

Tenglamaning ikkala tarafidan \frac{7}{2} ni ayirish.

$$x=1 x=-8$$