2x^2-5x+2xsqrt{x^2-5x+6}=3 eching

$$2 x ^ { 2 } - 5 x + 2 x \sqrt { x ^ { 2 } - 5 x + 6 } = 3$$

x uchun yechish

$$x=3$$

Yechim qadamlari

Tenglamaning ikkala tarafidan 2x^{2}-5x ni ayirish.

$$2x\sqrt{x^{2}-5x+6}=3-\left(2x^{2}-5x\right)$$

2x^{2}-5x teskarisini topish uchun har birining teskarisini toping.

$$2x\sqrt{x^{2}-5x+6}=3-2x^{2}+5x$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(2x\sqrt{x^{2}-5x+6}\right)^{2}=\left(3-2x^{2}+5x\right)^{2}$$

\left(2x\sqrt{x^{2}-5x+6}\right)^{2} ni kengaytirish.

$$2^{2}x^{2}\left(\sqrt{x^{2}-5x+6}\right)^{2}=\left(3-2x^{2}+5x\right)^{2}$$

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

$$4x^{2}\left(\sqrt{x^{2}-5x+6}\right)^{2}=\left(3-2x^{2}+5x\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x^{2}-5x+6} ga hisoblang va x^{2}-5x+6 ni qiymatni oling.

$$4x^{2}\left(x^{2}-5x+6\right)=\left(3-2x^{2}+5x\right)^{2}$$

4x^{2} ga x^{2}-5x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4x^{4}-20x^{3}+24x^{2}=\left(3-2x^{2}+5x\right)^{2}$$

3-2x^{2}+5x kvadratini chiqarish.

$$4x^{4}-20x^{3}+24x^{2}=4x^{4}-20x^{3}+13x^{2}+30x+9$$

Ikkala tarafdan 4x^{4} ni ayirish.

$$4x^{4}-20x^{3}+24x^{2}-4x^{4}=-20x^{3}+13x^{2}+30x+9$$

0 ni olish uchun 4x^{4} va -4x^{4} ni birlashtirish.

$$-20x^{3}+24x^{2}=-20x^{3}+13x^{2}+30x+9$$

20x^{3} ni ikki tarafga qo’shing.

$$-20x^{3}+24x^{2}+20x^{3}=13x^{2}+30x+9$$

0 ni olish uchun -20x^{3} va 20x^{3} ni birlashtirish.

$$24x^{2}=13x^{2}+30x+9$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$24x^{2}-13x^{2}=30x+9$$

11x^{2} ni olish uchun 24x^{2} va -13x^{2} ni birlashtirish.

$$11x^{2}=30x+9$$

Ikkala tarafdan 30x ni ayirish.

$$11x^{2}-30x=9$$

Ikkala tarafdan 9 ni ayirish.

$$11x^{2}-30x-9=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 11x^{2}+ax+bx-9 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-30 ab=11\left(-9\right)=-99$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -99-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-99 3,-33 9,-11$$

Har bir juftlik yigʻindisini hisoblang.

$$1-99=-98 3-33=-30 9-11=-2$$

Yechim – -30 yigʻindisini beruvchi juftlik.

$$a=-33 b=3$$

11x^{2}-30x-9 ni \left(11x^{2}-33x\right)+\left(3x-9\right) sifatida qaytadan yozish.

$$\left(11x^{2}-33x\right)+\left(3x-9\right)$$

Birinchi guruhda 11x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$11x\left(x-3\right)+3\left(x-3\right)$$

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(11x+3\right)$$

Tenglamani yechish uchun x-3=0 va 11x+3=0 ni yeching.

$$x=3 x=-\frac{3}{11}$$

2x^{2}-5x+2x\sqrt{x^{2}-5x+6}=3 tenglamasida x uchun 3 ni almashtiring.

$$2\times 3^{2}-5\times 3+2\times 3\sqrt{3^{2}-5\times 3+6}=3$$

Qisqartirish. x=3 tenglamani qoniqtiradi.

$$3=3$$

2x^{2}-5x+2x\sqrt{x^{2}-5x+6}=3 tenglamasida x uchun -\frac{3}{11} ni almashtiring.

$$2\left(-\frac{3}{11}\right)^{2}-5\left(-\frac{3}{11}\right)+2\left(-\frac{3}{11}\right)\sqrt{\left(-\frac{3}{11}\right)^{2}-5\left(-\frac{3}{11}\right)+6}=3$$

Qisqartirish. x=-\frac{3}{11} qiymati bu tenglamani qoniqtirmaydi.

$$\frac{3}{121}=3$$

2x\sqrt{x^{2}-5x+6}=3+5x-2x^{2} tenglamasi noyob yechimga ega.

$$x=3$$