(5x-2)^2-(3x-3)^2=0 eching

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

x uchun yechish

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

$$x=\frac{5}{8}=0,625$$

Guruhlash orqali faktorlash qadamlari

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

$$25x^{2}-20x+4-\left(3x-3\right)^{2}=0$$

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

$$25x^{2}-20x+4-\left(9x^{2}-18x+9\right)=0$$

9x^{2}-18x+9 teskarisini topish uchun har birining teskarisini toping.

$$25x^{2}-20x+4-9x^{2}+18x-9=0$$

16x^{2} ni olish uchun 25x^{2} va -9x^{2} ni birlashtirish.

$$16x^{2}-20x+4+18x-9=0$$

-2x ni olish uchun -20x va 18x ni birlashtirish.

$$16x^{2}-2x+4-9=0$$

-5 olish uchun 4 dan 9 ni ayirish.

$$16x^{2}-2x-5=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 16x^{2}+ax+bx-5 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-2 ab=16\left(-5\right)=-80$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -80-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

Yechim – -2 yigʻindisini beruvchi juftlik.

$$a=-10 b=8$$

16x^{2}-2x-5 ni \left(16x^{2}-10x\right)+\left(8x-5\right) sifatida qaytadan yozish.

$$\left(16x^{2}-10x\right)+\left(8x-5\right)$$

16x^{2}-10x ichida 2x ni ajrating.

$$2x\left(8x-5\right)+8x-5$$

Distributiv funktsiyasidan foydalangan holda 8x-5 umumiy terminini chiqaring.

$$\left(8x-5\right)\left(2x+1\right)$$

Tenglamani yechish uchun 8x-5=0 va 2x+1=0 ni yeching.

$$x=\frac{5}{8} x=-\frac{1}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$25x^{2}-20x+4-\left(3x-3\right)^{2}=0$$

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

$$25x^{2}-20x+4-\left(9x^{2}-18x+9\right)=0$$

9x^{2}-18x+9 teskarisini topish uchun har birining teskarisini toping.

$$25x^{2}-20x+4-9x^{2}+18x-9=0$$

16x^{2} ni olish uchun 25x^{2} va -9x^{2} ni birlashtirish.

$$16x^{2}-20x+4+18x-9=0$$

-2x ni olish uchun -20x va 18x ni birlashtirish.

$$16x^{2}-2x+4-9=0$$

-5 olish uchun 4 dan 9 ni ayirish.

$$16x^{2}-2x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 16 ni a, -2 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 16\left(-5\right)}}{2\times 16}$$

-2 kvadratini chiqarish.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 16\left(-5\right)}}{2\times 16}$$

-4 ni 16 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-64\left(-5\right)}}{2\times 16}$$

-64 ni -5 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+320}}{2\times 16}$$

4 ni 320 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{324}}{2\times 16}$$

324 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±18}{2\times 16}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±18}{2\times 16}$$

2 ni 16 marotabaga ko'paytirish.

$$x=\frac{2±18}{32}$$

x=\frac{2±18}{32} tenglamasini yeching, bunda ± musbat. 2 ni 18 ga qo'shish.

$$x=\frac{20}{32}$$

\frac{20}{32} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{2±18}{32} tenglamasini yeching, bunda ± manfiy. 2 dan 18 ni ayirish.

$$x=-\frac{16}{32}$$

\frac{-16}{32} ulushini 16 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=\frac{5}{8} x=-\frac{1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$25x^{2}-20x+4-\left(3x-3\right)^{2}=0$$

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

$$25x^{2}-20x+4-\left(9x^{2}-18x+9\right)=0$$

9x^{2}-18x+9 teskarisini topish uchun har birining teskarisini toping.

$$25x^{2}-20x+4-9x^{2}+18x-9=0$$

16x^{2} ni olish uchun 25x^{2} va -9x^{2} ni birlashtirish.

$$16x^{2}-20x+4+18x-9=0$$

-2x ni olish uchun -20x va 18x ni birlashtirish.

$$16x^{2}-2x+4-9=0$$

-5 olish uchun 4 dan 9 ni ayirish.

$$16x^{2}-2x-5=0$$

5 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$16x^{2}-2x=5$$

Ikki tarafini 16 ga bo‘ling.

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

16 ga bo'lish 16 ga ko'paytirishni bekor qiladi.

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

\frac{-2}{16} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}-\frac{1}{8}x=\frac{5}{16}$$

-\frac{1}{8} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{16} olish uchun. Keyin, -\frac{1}{16} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\frac{5}{16}+\left(-\frac{1}{16}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{16} kvadratini chiqarish.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{5}{16}+\frac{1}{256}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{16} ni \frac{1}{256} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{81}{256}$$

x^{2}-\frac{1}{8}x+\frac{1}{256} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{81}{256}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{81}{256}}$$

Qisqartirish.

$$x-\frac{1}{16}=\frac{9}{16} x-\frac{1}{16}=-\frac{9}{16}$$

\frac{1}{16} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{5}{8} x=-\frac{1}{2}$$