(3x-2)^2=(x-1)^2 eching

$$( 3 x - 2 ) ^ { 2 } = ( x - 1 ) ^ { 2 }$$

x uchun yechish

$$x=\frac{1}{2}=0,5$$

$$x=\frac{3}{4}=0,75$$

Guruhlash orqali faktorlash qadamlari

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

$$9x^{2}-12x+4=\left(x-1\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}-2x+1$$

Ikkala tarafdan x^{2} ni ayirish.

$$9x^{2}-12x+4-x^{2}=-2x+1$$

8x^{2} ni olish uchun 9x^{2} va -x^{2} ni birlashtirish.

$$8x^{2}-12x+4=-2x+1$$

2x ni ikki tarafga qo’shing.

$$8x^{2}-12x+4+2x=1$$

-10x ni olish uchun -12x va 2x ni birlashtirish.

$$8x^{2}-10x+4=1$$

Ikkala tarafdan 1 ni ayirish.

$$8x^{2}-10x+4-1=0$$

3 olish uchun 4 dan 1 ni ayirish.

$$8x^{2}-10x+3=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 8x^{2}+ax+bx+3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-10 ab=8\times 3=24$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 24-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

Yechim – -10 yigʻindisini beruvchi juftlik.

$$a=-6 b=-4$$

8x^{2}-10x+3 ni \left(8x^{2}-6x\right)+\left(-4x+3\right) sifatida qaytadan yozish.

$$\left(8x^{2}-6x\right)+\left(-4x+3\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$2x\left(4x-3\right)-\left(4x-3\right)$$

Distributiv funktsiyasidan foydalangan holda 4x-3 umumiy terminini chiqaring.

$$\left(4x-3\right)\left(2x-1\right)$$

Tenglamani yechish uchun 4x-3=0 va 2x-1=0 ni yeching.

$$x=\frac{3}{4} x=\frac{1}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$9x^{2}-12x+4=\left(x-1\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}-2x+1$$

Ikkala tarafdan x^{2} ni ayirish.

$$9x^{2}-12x+4-x^{2}=-2x+1$$

8x^{2} ni olish uchun 9x^{2} va -x^{2} ni birlashtirish.

$$8x^{2}-12x+4=-2x+1$$

2x ni ikki tarafga qo’shing.

$$8x^{2}-12x+4+2x=1$$

-10x ni olish uchun -12x va 2x ni birlashtirish.

$$8x^{2}-10x+4=1$$

Ikkala tarafdan 1 ni ayirish.

$$8x^{2}-10x+4-1=0$$

3 olish uchun 4 dan 1 ni ayirish.

$$8x^{2}-10x+3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 8 ni a, -10 ni b va 3 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 8\times 3}}{2\times 8}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 8\times 3}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-32\times 3}}{2\times 8}$$

-32 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 8}$$

100 ni -96 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 8}$$

4 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2}{2\times 8}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{10±2}{16}$$

x=\frac{10±2}{16} tenglamasini yeching, bunda ± musbat. 10 ni 2 ga qo'shish.

$$x=\frac{12}{16}$$

\frac{12}{16} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{10±2}{16} tenglamasini yeching, bunda ± manfiy. 10 dan 2 ni ayirish.

$$x=\frac{8}{16}$$

\frac{8}{16} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=\frac{3}{4} x=\frac{1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$9x^{2}-12x+4=\left(x-1\right)^{2}$$

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

$$9x^{2}-12x+4=x^{2}-2x+1$$

Ikkala tarafdan x^{2} ni ayirish.

$$9x^{2}-12x+4-x^{2}=-2x+1$$

8x^{2} ni olish uchun 9x^{2} va -x^{2} ni birlashtirish.

$$8x^{2}-12x+4=-2x+1$$

2x ni ikki tarafga qo’shing.

$$8x^{2}-12x+4+2x=1$$

-10x ni olish uchun -12x va 2x ni birlashtirish.

$$8x^{2}-10x+4=1$$

Ikkala tarafdan 4 ni ayirish.

$$8x^{2}-10x=1-4$$

-3 olish uchun 1 dan 4 ni ayirish.

$$8x^{2}-10x=-3$$

Ikki tarafini 8 ga bo‘ling.

$$\frac{8x^{2}-10x}{8}=-\frac{3}{8}$$

8 ga bo'lish 8 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{10}{8}\right)x=-\frac{3}{8}$$

\frac{-10}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{5}{4}x=-\frac{3}{8}$$

-\frac{5}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{8} olish uchun. Keyin, -\frac{5}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=-\frac{3}{8}+\left(-\frac{5}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{8} kvadratini chiqarish.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{3}{8}+\frac{25}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{8} ni \frac{25}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{1}{64}$$

x^{2}-\frac{5}{4}x+\frac{25}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{1}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Qisqartirish.

$$x-\frac{5}{8}=\frac{1}{8} x-\frac{5}{8}=-\frac{1}{8}$$

\frac{5}{8} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3}{4} x=\frac{1}{2}$$