(5x-2)^2-(2x-1)(2x+1)=47+x eching

$$( 5 x - 2 ) ^ { 2 } - ( 2 x - 1 ) ( 2 x + 1 ) = 47 + x$$

x uchun yechish

$$x=-1$$

$$x=2$$

Guruhlash orqali faktorlash qadamlari

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

$$25x^{2}-20x+4-\left(2x-1\right)\left(2x+1\right)=47+x$$

Hisoblang: \left(2x-1\right)\left(2x+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.

$$25x^{2}-20x+4-\left(\left(2x\right)^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(2^{2}x^{2}-1\right)=47+x$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$25x^{2}-20x+4-\left(4x^{2}-1\right)=47+x$$

4x^{2}-1 teskarisini topish uchun har birining teskarisini toping.

$$25x^{2}-20x+4-4x^{2}+1=47+x$$

21x^{2} ni olish uchun 25x^{2} va -4x^{2} ni birlashtirish.

$$21x^{2}-20x+4+1=47+x$$

5 olish uchun 4 va 1'ni qo'shing.

$$21x^{2}-20x+5=47+x$$

Ikkala tarafdan 47 ni ayirish.

$$21x^{2}-20x+5-47=x$$

-42 olish uchun 5 dan 47 ni ayirish.

$$21x^{2}-20x-42=x$$

Ikkala tarafdan x ni ayirish.

$$21x^{2}-20x-42-x=0$$

-21x ni olish uchun -20x va -x ni birlashtirish.

$$21x^{2}-21x-42=0$$

Ikki tarafini 21 ga bo‘ling.

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

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

$$a+b=-1 ab=1\left(-2\right)=-2$$

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.

$$a=-2 b=1$$

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

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

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

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

Distributiv funktsiyasidan foydalangan holda x-2 umumiy terminini chiqaring.

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

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

$$x=2 x=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$25x^{2}-20x+4-\left(2x-1\right)\left(2x+1\right)=47+x$$

Hisoblang: \left(2x-1\right)\left(2x+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.

$$25x^{2}-20x+4-\left(\left(2x\right)^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(2^{2}x^{2}-1\right)=47+x$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$25x^{2}-20x+4-\left(4x^{2}-1\right)=47+x$$

4x^{2}-1 teskarisini topish uchun har birining teskarisini toping.

$$25x^{2}-20x+4-4x^{2}+1=47+x$$

21x^{2} ni olish uchun 25x^{2} va -4x^{2} ni birlashtirish.

$$21x^{2}-20x+4+1=47+x$$

5 olish uchun 4 va 1'ni qo'shing.

$$21x^{2}-20x+5=47+x$$

Ikkala tarafdan 47 ni ayirish.

$$21x^{2}-20x+5-47=x$$

-42 olish uchun 5 dan 47 ni ayirish.

$$21x^{2}-20x-42=x$$

Ikkala tarafdan x ni ayirish.

$$21x^{2}-20x-42-x=0$$

-21x ni olish uchun -20x va -x ni birlashtirish.

$$21x^{2}-21x-42=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 21 ni a, -21 ni b va -42 ni c bilan almashtiring.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 21\left(-42\right)}}{2\times 21}$$

-21 kvadratini chiqarish.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 21\left(-42\right)}}{2\times 21}$$

-4 ni 21 marotabaga ko'paytirish.

$$x=\frac{-\left(-21\right)±\sqrt{441-84\left(-42\right)}}{2\times 21}$$

-84 ni -42 marotabaga ko'paytirish.

$$x=\frac{-\left(-21\right)±\sqrt{441+3528}}{2\times 21}$$

441 ni 3528 ga qo'shish.

$$x=\frac{-\left(-21\right)±\sqrt{3969}}{2\times 21}$$

3969 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-21\right)±63}{2\times 21}$$

-21 ning teskarisi 21 ga teng.

$$x=\frac{21±63}{2\times 21}$$

2 ni 21 marotabaga ko'paytirish.

$$x=\frac{21±63}{42}$$

x=\frac{21±63}{42} tenglamasini yeching, bunda ± musbat. 21 ni 63 ga qo'shish.

$$x=\frac{84}{42}$$

84 ni 42 ga bo'lish.

$$x=2$$

x=\frac{21±63}{42} tenglamasini yeching, bunda ± manfiy. 21 dan 63 ni ayirish.

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

-42 ni 42 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

$$x=2 x=-1$$

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.

$$25x^{2}-20x+4-\left(2x-1\right)\left(2x+1\right)=47+x$$

Hisoblang: \left(2x-1\right)\left(2x+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.

$$25x^{2}-20x+4-\left(\left(2x\right)^{2}-1\right)=47+x$$

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

$$25x^{2}-20x+4-\left(2^{2}x^{2}-1\right)=47+x$$

2 daraja ko‘rsatkichini 2 ga hisoblang va 4 ni qiymatni oling.

$$25x^{2}-20x+4-\left(4x^{2}-1\right)=47+x$$

4x^{2}-1 teskarisini topish uchun har birining teskarisini toping.

$$25x^{2}-20x+4-4x^{2}+1=47+x$$

21x^{2} ni olish uchun 25x^{2} va -4x^{2} ni birlashtirish.

$$21x^{2}-20x+4+1=47+x$$

5 olish uchun 4 va 1'ni qo'shing.

$$21x^{2}-20x+5=47+x$$

Ikkala tarafdan x ni ayirish.

$$21x^{2}-20x+5-x=47$$

-21x ni olish uchun -20x va -x ni birlashtirish.

$$21x^{2}-21x+5=47$$

Ikkala tarafdan 5 ni ayirish.

$$21x^{2}-21x=47-5$$

42 olish uchun 47 dan 5 ni ayirish.

$$21x^{2}-21x=42$$

Ikki tarafini 21 ga bo‘ling.

$$\frac{21x^{2}-21x}{21}=\frac{42}{21}$$

21 ga bo'lish 21 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{21}{21}\right)x=\frac{42}{21}$$

-21 ni 21 ga bo'lish.

$$x^{2}-x=\frac{42}{21}$$

42 ni 21 ga bo'lish.

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{2} kvadratini chiqarish.

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

2 ni \frac{1}{4} ga qo'shish.

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

x^{2}-x+\frac{1}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

\frac{1}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=2 x=-1$$