sqrt{x+3}+sqrt{x+6}=sqrt{x+11} eching

$$\sqrt { x + 3 } + \sqrt { x + 6 } = \sqrt { x + 11 }$$

x uchun yechish

$$x=-2$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{x+3}+\sqrt{x+6}\right)^{2} kengaytirilishi uchun ishlating.

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

2 daraja ko‘rsatkichini \sqrt{x+3} ga hisoblang va x+3 ni qiymatni oling.

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

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

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

2x ni olish uchun x va x ni birlashtirish.

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

9 olish uchun 3 va 6'ni qo'shing.

$$2x+9+2\sqrt{x+3}\sqrt{x+6}=\left(\sqrt{x+11}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x+11} ga hisoblang va x+11 ni qiymatni oling.

$$2x+9+2\sqrt{x+3}\sqrt{x+6}=x+11$$

Tenglamaning ikkala tarafidan 2x+9 ni ayirish.

$$2\sqrt{x+3}\sqrt{x+6}=x+11-\left(2x+9\right)$$

2x+9 teskarisini topish uchun har birining teskarisini toping.

$$2\sqrt{x+3}\sqrt{x+6}=x+11-2x-9$$

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

$$2\sqrt{x+3}\sqrt{x+6}=-x+11-9$$

2 olish uchun 11 dan 9 ni ayirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\left(2\sqrt{x+3}\sqrt{x+6}\right)^{2} ni kengaytirish.

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

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

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

2 daraja ko‘rsatkichini \sqrt{x+3} ga hisoblang va x+3 ni qiymatni oling.

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

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

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

4 ga x+3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(4x+12\right)\left(x+6\right)=\left(-x+2\right)^{2}$$

4x+12 ifodaning har bir elementini x+6 ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$4x^{2}+24x+12x+72=\left(-x+2\right)^{2}$$

36x ni olish uchun 24x va 12x ni birlashtirish.

$$4x^{2}+36x+72=\left(-x+2\right)^{2}$$

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

$$4x^{2}+36x+72=x^{2}-4x+4$$

Ikkala tarafdan x^{2} ni ayirish.

$$4x^{2}+36x+72-x^{2}=-4x+4$$

3x^{2} ni olish uchun 4x^{2} va -x^{2} ni birlashtirish.

$$3x^{2}+36x+72=-4x+4$$

4x ni ikki tarafga qo’shing.

$$3x^{2}+36x+72+4x=4$$

40x ni olish uchun 36x va 4x ni birlashtirish.

$$3x^{2}+40x+72=4$$

Ikkala tarafdan 4 ni ayirish.

$$3x^{2}+40x+72-4=0$$

68 olish uchun 72 dan 4 ni ayirish.

$$3x^{2}+40x+68=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 3x^{2}+ax+bx+68 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=40 ab=3\times 68=204$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 204-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,204 2,102 3,68 4,51 6,34 12,17$$

Har bir juftlik yigʻindisini hisoblang.

$$1+204=205 2+102=104 3+68=71 4+51=55 6+34=40 12+17=29$$

Yechim – 40 yigʻindisini beruvchi juftlik.

$$a=6 b=34$$

3x^{2}+40x+68 ni \left(3x^{2}+6x\right)+\left(34x+68\right) sifatida qaytadan yozish.

$$\left(3x^{2}+6x\right)+\left(34x+68\right)$$

Birinchi guruhda 3x ni va ikkinchi guruhda 34 ni faktordan chiqaring.

$$3x\left(x+2\right)+34\left(x+2\right)$$

Distributiv funktsiyasidan foydalangan holda x+2 umumiy terminini chiqaring.

$$\left(x+2\right)\left(3x+34\right)$$

Tenglamani yechish uchun x+2=0 va 3x+34=0 ni yeching.

$$x=-2 x=-\frac{34}{3}$$

\sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11} tenglamasida x uchun -\frac{34}{3} ni almashtiring. \sqrt{-\frac{34}{3}+3} ifodasi noaniq, chunki ildiz ostidagi qiymat manfiy boʻlishi mumkin emas.

$$\sqrt{-\frac{34}{3}+3}+\sqrt{-\frac{34}{3}+6}=\sqrt{-\frac{34}{3}+11}$$

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

$$\sqrt{-2+3}+\sqrt{-2+6}=\sqrt{-2+11}$$

Qisqartirish. x=-2 tenglamani qoniqtiradi.

$$3=3$$

\sqrt{x+3}+\sqrt{x+6}=\sqrt{x+11} tenglamasi noyob yechimga ega.

$$x=-2$$