(sqrt{x+2}+sqrt{18-x})^2=(sqrt{10+13x})^2 eching

$$( \sqrt { x + 2 } + \sqrt { 18 - x } ) ^ { 2 } = ( \sqrt { 10 + 13 x } ) ^ { 2 }$$

x uchun yechish

$$x=2$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{x+2}+\sqrt{18-x}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(\sqrt{x+2}\right)^{2}+2\sqrt{x+2}\sqrt{18-x}+\left(\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2}$$

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

$$x+2+2\sqrt{x+2}\sqrt{18-x}+\left(\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{18-x} ga hisoblang va 18-x ni qiymatni oling.

$$x+2+2\sqrt{x+2}\sqrt{18-x}+18-x=\left(\sqrt{10+13x}\right)^{2}$$

20 olish uchun 2 va 18'ni qo'shing.

$$x+20+2\sqrt{x+2}\sqrt{18-x}-x=\left(\sqrt{10+13x}\right)^{2}$$

0 ni olish uchun x va -x ni birlashtirish.

$$20+2\sqrt{x+2}\sqrt{18-x}=\left(\sqrt{10+13x}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{10+13x} ga hisoblang va 10+13x ni qiymatni oling.

$$20+2\sqrt{x+2}\sqrt{18-x}=10+13x$$

Ikkala tarafdan 13x ni ayirish.

$$20+2\sqrt{x+2}\sqrt{18-x}-13x=10$$

Ikkala tarafdan 20 ni ayirish.

$$2\sqrt{x+2}\sqrt{18-x}-13x=10-20$$

-10 olish uchun 10 dan 20 ni ayirish.

$$2\sqrt{x+2}\sqrt{18-x}-13x=-10$$

Tenglamaning ikkala tarafidan -13x ni ayirish.

$$2\sqrt{x+2}\sqrt{18-x}=-10+13x$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(2\sqrt{x+2}\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

\left(2\sqrt{x+2}\sqrt{18-x}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{x+2}\right)^{2}\left(\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

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

$$4\left(\sqrt{x+2}\right)^{2}\left(\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

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

$$4\left(x+2\right)\left(\sqrt{18-x}\right)^{2}=\left(13x-10\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{18-x} ga hisoblang va 18-x ni qiymatni oling.

$$4\left(x+2\right)\left(18-x\right)=\left(13x-10\right)^{2}$$

4 ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(4x+8\right)\left(18-x\right)=\left(13x-10\right)^{2}$$

4x+8 ga 18-x ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$64x-4x^{2}+144=\left(13x-10\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(13x-10\right)^{2} kengaytirilishi uchun ishlating.

$$64x-4x^{2}+144=169x^{2}-260x+100$$

Ikkala tarafdan 169x^{2} ni ayirish.

$$64x-4x^{2}+144-169x^{2}=-260x+100$$

-173x^{2} ni olish uchun -4x^{2} va -169x^{2} ni birlashtirish.

$$64x-173x^{2}+144=-260x+100$$

260x ni ikki tarafga qo’shing.

$$64x-173x^{2}+144+260x=100$$

324x ni olish uchun 64x va 260x ni birlashtirish.

$$324x-173x^{2}+144=100$$

Ikkala tarafdan 100 ni ayirish.

$$324x-173x^{2}+144-100=0$$

44 olish uchun 144 dan 100 ni ayirish.

$$324x-173x^{2}+44=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$-173x^{2}+324x+44=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -173x^{2}+ax+bx+44 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=324 ab=-173\times 44=-7612$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -7612-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,7612 -2,3806 -4,1903 -11,692 -22,346 -44,173$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+7612=7611 -2+3806=3804 -4+1903=1899 -11+692=681 -22+346=324 -44+173=129$$

Yechim – 324 yigʻindisini beruvchi juftlik.

$$a=346 b=-22$$

-173x^{2}+324x+44 ni \left(-173x^{2}+346x\right)+\left(-22x+44\right) sifatida qaytadan yozish.

$$\left(-173x^{2}+346x\right)+\left(-22x+44\right)$$

Birinchi guruhda 173x ni va ikkinchi guruhda 22 ni faktordan chiqaring.

$$173x\left(-x+2\right)+22\left(-x+2\right)$$

Distributiv funktsiyasidan foydalangan holda -x+2 umumiy terminini chiqaring.

$$\left(-x+2\right)\left(173x+22\right)$$

Tenglamani yechish uchun -x+2=0 va 173x+22=0 ni yeching.

$$x=2 x=-\frac{22}{173}$$

\left(\sqrt{x+2}+\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2} tenglamasida x uchun 2 ni almashtiring.

$$\left(\sqrt{2+2}+\sqrt{18-2}\right)^{2}=\left(\sqrt{10+13\times 2}\right)^{2}$$

Qisqartirish. x=2 tenglamani qoniqtiradi.

$$36=36$$

\left(\sqrt{x+2}+\sqrt{18-x}\right)^{2}=\left(\sqrt{10+13x}\right)^{2} tenglamasida x uchun -\frac{22}{173} ni almashtiring.

$$\left(\sqrt{-\frac{22}{173}+2}+\sqrt{18-\left(-\frac{22}{173}\right)}\right)^{2}=\left(\sqrt{10+13\left(-\frac{22}{173}\right)}\right)^{2}$$

Qisqartirish. x=-\frac{22}{173} qiymati bu tenglamani qoniqtirmaydi.

$$\frac{5476}{173}=\frac{1444}{173}$$

2\sqrt{18-x}\sqrt{x+2}=13x-10 tenglamasi noyob yechimga ega.

$$x=2$$