sqrt{x+6}-sqrt{9x+70}=-2sqrt{x+9} eching

$$\sqrt { x + 6 } - \sqrt { 9 x + 70 } = - 2 \sqrt { x + 9 }$$

x uchun yechish

$$x=-5$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{x+6}-\sqrt{9x+70}\right)^{2}=\left(-2\sqrt{x+9}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{x+6}-\sqrt{9x+70}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(\sqrt{x+6}\right)^{2}-2\sqrt{x+6}\sqrt{9x+70}+\left(\sqrt{9x+70}\right)^{2}=\left(-2\sqrt{x+9}\right)^{2}$$

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

$$x+6-2\sqrt{x+6}\sqrt{9x+70}+\left(\sqrt{9x+70}\right)^{2}=\left(-2\sqrt{x+9}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{9x+70} ga hisoblang va 9x+70 ni qiymatni oling.

$$x+6-2\sqrt{x+6}\sqrt{9x+70}+9x+70=\left(-2\sqrt{x+9}\right)^{2}$$

10x ni olish uchun x va 9x ni birlashtirish.

$$10x+6-2\sqrt{x+6}\sqrt{9x+70}+70=\left(-2\sqrt{x+9}\right)^{2}$$

76 olish uchun 6 va 70'ni qo'shing.

$$10x+76-2\sqrt{x+6}\sqrt{9x+70}=\left(-2\sqrt{x+9}\right)^{2}$$

\left(-2\sqrt{x+9}\right)^{2} ni kengaytirish.

$$10x+76-2\sqrt{x+6}\sqrt{9x+70}=\left(-2\right)^{2}\left(\sqrt{x+9}\right)^{2}$$

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

$$10x+76-2\sqrt{x+6}\sqrt{9x+70}=4\left(\sqrt{x+9}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x+9} ga hisoblang va x+9 ni qiymatni oling.

$$10x+76-2\sqrt{x+6}\sqrt{9x+70}=4\left(x+9\right)$$

4 ga x+9 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10x+76-2\sqrt{x+6}\sqrt{9x+70}=4x+36$$

Tenglamaning ikkala tarafidan 10x+76 ni ayirish.

$$-2\sqrt{x+6}\sqrt{9x+70}=4x+36-\left(10x+76\right)$$

10x+76 teskarisini topish uchun har birining teskarisini toping.

$$-2\sqrt{x+6}\sqrt{9x+70}=4x+36-10x-76$$

-6x ni olish uchun 4x va -10x ni birlashtirish.

$$-2\sqrt{x+6}\sqrt{9x+70}=-6x+36-76$$

-40 olish uchun 36 dan 76 ni ayirish.

$$-2\sqrt{x+6}\sqrt{9x+70}=-6x-40$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(-2\sqrt{x+6}\sqrt{9x+70}\right)^{2}=\left(-6x-40\right)^{2}$$

\left(-2\sqrt{x+6}\sqrt{9x+70}\right)^{2} ni kengaytirish.

$$\left(-2\right)^{2}\left(\sqrt{x+6}\right)^{2}\left(\sqrt{9x+70}\right)^{2}=\left(-6x-40\right)^{2}$$

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

$$4\left(\sqrt{x+6}\right)^{2}\left(\sqrt{9x+70}\right)^{2}=\left(-6x-40\right)^{2}$$

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

$$4\left(x+6\right)\left(\sqrt{9x+70}\right)^{2}=\left(-6x-40\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{9x+70} ga hisoblang va 9x+70 ni qiymatni oling.

$$4\left(x+6\right)\left(9x+70\right)=\left(-6x-40\right)^{2}$$

4 ga x+6 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(4x+24\right)\left(9x+70\right)=\left(-6x-40\right)^{2}$$

4x+24 ifodaning har bir elementini 9x+70 ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$36x^{2}+280x+216x+1680=\left(-6x-40\right)^{2}$$

496x ni olish uchun 280x va 216x ni birlashtirish.

$$36x^{2}+496x+1680=\left(-6x-40\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(-6x-40\right)^{2} kengaytirilishi uchun ishlating.

$$36x^{2}+496x+1680=36x^{2}+480x+1600$$

Ikkala tarafdan 36x^{2} ni ayirish.

$$36x^{2}+496x+1680-36x^{2}=480x+1600$$

0 ni olish uchun 36x^{2} va -36x^{2} ni birlashtirish.

$$496x+1680=480x+1600$$

Ikkala tarafdan 480x ni ayirish.

$$496x+1680-480x=1600$$

16x ni olish uchun 496x va -480x ni birlashtirish.

$$16x+1680=1600$$

Ikkala tarafdan 1680 ni ayirish.

$$16x=1600-1680$$

-80 olish uchun 1600 dan 1680 ni ayirish.

$$16x=-80$$

Ikki tarafini 16 ga bo‘ling.

$$x=\frac{-80}{16}$$

-5 ni olish uchun -80 ni 16 ga bo‘ling.

$$x=-5$$

\sqrt{x+6}-\sqrt{9x+70}=-2\sqrt{x+9} tenglamasida x uchun -5 ni almashtiring.

$$\sqrt{-5+6}-\sqrt{9\left(-5\right)+70}=-2\sqrt{-5+9}$$

Qisqartirish. x=-5 tenglamani qoniqtiradi.

$$-4=-4$$

\sqrt{x+6}-\sqrt{9x+70}=-2\sqrt{x+9} tenglamasi noyob yechimga ega.

$$x=-5$$