left(3x-4right)^2=9x-12 eching

$${ \left(3x-4 \right) }^{ 2 } = 9x-12$$

x uchun yechish

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1,333333333$$

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2,333333333$$

Guruhlash orqali faktorlash qadamlari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3x-4\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}-24x+16=9x-12$$

Ikkala tarafdan 9x ni ayirish.

$$9x^{2}-24x+16-9x=-12$$

-33x ni olish uchun -24x va -9x ni birlashtirish.

$$9x^{2}-33x+16=-12$$

12 ni ikki tarafga qo’shing.

$$9x^{2}-33x+16+12=0$$

28 olish uchun 16 va 12'ni qo'shing.

$$9x^{2}-33x+28=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 9x^{2}+ax+bx+28 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-33 ab=9\times 28=252$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 252-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-252 -2,-126 -3,-84 -4,-63 -6,-42 -7,-36 -9,-28 -12,-21 -14,-18$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-252=-253 -2-126=-128 -3-84=-87 -4-63=-67 -6-42=-48 -7-36=-43 -9-28=-37 -12-21=-33 -14-18=-32$$

Yechim – -33 yigʻindisini beruvchi juftlik.

$$a=-21 b=-12$$

9x^{2}-33x+28 ni \left(9x^{2}-21x\right)+\left(-12x+28\right) sifatida qaytadan yozish.

$$\left(9x^{2}-21x\right)+\left(-12x+28\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda -4 ni faktordan chiqaring.

$$3x\left(3x-7\right)-4\left(3x-7\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-7 umumiy terminini chiqaring.

$$\left(3x-7\right)\left(3x-4\right)$$

Tenglamani yechish uchun 3x-7=0 va 3x-4=0 ni yeching.

$$x=\frac{7}{3} x=\frac{4}{3}$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3x-4\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}-24x+16=9x-12$$

Ikkala tarafdan 9x ni ayirish.

$$9x^{2}-24x+16-9x=-12$$

-33x ni olish uchun -24x va -9x ni birlashtirish.

$$9x^{2}-33x+16=-12$$

12 ni ikki tarafga qo’shing.

$$9x^{2}-33x+16+12=0$$

28 olish uchun 16 va 12'ni qo'shing.

$$9x^{2}-33x+28=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, -33 ni b va 28 ni c bilan almashtiring.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 9\times 28}}{2\times 9}$$

-33 kvadratini chiqarish.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\times 9\times 28}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-33\right)±\sqrt{1089-36\times 28}}{2\times 9}$$

-36 ni 28 marotabaga ko'paytirish.

$$x=\frac{-\left(-33\right)±\sqrt{1089-1008}}{2\times 9}$$

1089 ni -1008 ga qo'shish.

$$x=\frac{-\left(-33\right)±\sqrt{81}}{2\times 9}$$

81 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-33\right)±9}{2\times 9}$$

-33 ning teskarisi 33 ga teng.

$$x=\frac{33±9}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{33±9}{18}$$

x=\frac{33±9}{18} tenglamasini yeching, bunda ± musbat. 33 ni 9 ga qo'shish.

$$x=\frac{42}{18}$$

\frac{42}{18} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{7}{3}$$

x=\frac{33±9}{18} tenglamasini yeching, bunda ± manfiy. 33 dan 9 ni ayirish.

$$x=\frac{24}{18}$$

\frac{24}{18} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{4}{3}$$

Tenglama yechildi.

$$x=\frac{7}{3} x=\frac{4}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3x-4\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}-24x+16=9x-12$$

Ikkala tarafdan 9x ni ayirish.

$$9x^{2}-24x+16-9x=-12$$

-33x ni olish uchun -24x va -9x ni birlashtirish.

$$9x^{2}-33x+16=-12$$

Ikkala tarafdan 16 ni ayirish.

$$9x^{2}-33x=-12-16$$

-28 olish uchun -12 dan 16 ni ayirish.

$$9x^{2}-33x=-28$$

Ikki tarafini 9 ga bo‘ling.

$$\frac{9x^{2}-33x}{9}=-\frac{28}{9}$$

9 ga bo'lish 9 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{33}{9}\right)x=-\frac{28}{9}$$

\frac{-33}{9} ulushini 3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{11}{3}x=-\frac{28}{9}$$

-\frac{11}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{11}{6} olish uchun. Keyin, -\frac{11}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-\frac{28}{9}+\left(-\frac{11}{6}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{11}{6} kvadratini chiqarish.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{28}{9}+\frac{121}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{28}{9} ni \frac{121}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{1}{4}$$

x^{2}-\frac{11}{3}x+\frac{121}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{1}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Qisqartirish.

$$x-\frac{11}{6}=\frac{1}{2} x-\frac{11}{6}=-\frac{1}{2}$$

\frac{11}{6} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{7}{3} x=\frac{4}{3}$$