left(5x-2right)^2=9 eching

$${ \left(5x-2 \right) }^{ 2 } =9$$

x uchun yechish

$$x=1$$

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

Guruhlash orqali faktorlash qadamlari

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

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

Ikkala tarafdan 9 ni ayirish.

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

-5 olish uchun 4 dan 9 ni ayirish.

$$25x^{2}-20x-5=0$$

Ikki tarafini 5 ga bo‘ling.

$$5x^{2}-4x-1=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 5x^{2}+ax+bx-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-4 ab=5\left(-1\right)=-5$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-5 b=1$$

5x^{2}-4x-1 ni \left(5x^{2}-5x\right)+\left(x-1\right) sifatida qaytadan yozish.

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

5x^{2}-5x ichida 5x ni ajrating.

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

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

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

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

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

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

Ikkala tarafdan 9 ni ayirish.

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

-5 olish uchun 4 dan 9 ni ayirish.

$$25x^{2}-20x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, -20 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 25\left(-5\right)}}{2\times 25}$$

-20 kvadratini chiqarish.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 25\left(-5\right)}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400-100\left(-5\right)}}{2\times 25}$$

-100 ni -5 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400+500}}{2\times 25}$$

400 ni 500 ga qo'shish.

$$x=\frac{-\left(-20\right)±\sqrt{900}}{2\times 25}$$

900 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-20\right)±30}{2\times 25}$$

-20 ning teskarisi 20 ga teng.

$$x=\frac{20±30}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$x=\frac{20±30}{50}$$

x=\frac{20±30}{50} tenglamasini yeching, bunda ± musbat. 20 ni 30 ga qo'shish.

$$x=\frac{50}{50}$$

50 ni 50 ga bo'lish.

$$x=1$$

x=\frac{20±30}{50} tenglamasini yeching, bunda ± manfiy. 20 dan 30 ni ayirish.

$$x=-\frac{10}{50}$$

\frac{-10}{50} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

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

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.

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

Ikkala tarafdan 4 ni ayirish.

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

5 olish uchun 9 dan 4 ni ayirish.

$$25x^{2}-20x=5$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25x^{2}-20x}{25}=\frac{5}{25}$$

25 ga bo'lish 25 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{20}{25}\right)x=\frac{5}{25}$$

\frac{-20}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

\frac{5}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{2}{5} kvadratini chiqarish.

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{5} ni \frac{4}{25} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{9}{25}$$

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

$$\left(x-\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Qisqartirish.

$$x-\frac{2}{5}=\frac{3}{5} x-\frac{2}{5}=-\frac{3}{5}$$

\frac{2}{5} ni tenglamaning ikkala tarafiga qo'shish.

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