sqrt{x-4}-sqrt{4x-27}+sqrt{x-9}=0 eching

$$\sqrt{ x-4 } - \sqrt{ 4x-27 } + \sqrt{ x-9 } =0$$

x uchun yechish

$$x=13$$

Yechim qadamlari

Tenglamaning ikkala tarafidan -\sqrt{4x-27}+\sqrt{x-9} ni ayirish.

$$\sqrt{x-4}=-\left(-\sqrt{4x-27}+\sqrt{x-9}\right)$$

-\sqrt{4x-27}+\sqrt{x-9} teskarisini topish uchun har birining teskarisini toping.

$$\sqrt{x-4}=-\left(-\sqrt{4x-27}\right)-\sqrt{x-9}$$

-\sqrt{4x-27} ning teskarisi \sqrt{4x-27} ga teng.

$$\sqrt{x-4}=\sqrt{4x-27}-\sqrt{x-9}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{x-4}\right)^{2}=\left(\sqrt{4x-27}-\sqrt{x-9}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x-4} ga hisoblang va x-4 ni qiymatni oling.

$$x-4=\left(\sqrt{4x-27}-\sqrt{x-9}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{4x-27}-\sqrt{x-9}\right)^{2} kengaytirilishi uchun ishlating.

$$x-4=\left(\sqrt{4x-27}\right)^{2}-2\sqrt{4x-27}\sqrt{x-9}+\left(\sqrt{x-9}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{4x-27} ga hisoblang va 4x-27 ni qiymatni oling.

$$x-4=4x-27-2\sqrt{4x-27}\sqrt{x-9}+\left(\sqrt{x-9}\right)^{2}$$

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

$$x-4=4x-27-2\sqrt{4x-27}\sqrt{x-9}+x-9$$

5x ni olish uchun 4x va x ni birlashtirish.

$$x-4=5x-27-2\sqrt{4x-27}\sqrt{x-9}-9$$

-36 olish uchun -27 dan 9 ni ayirish.

$$x-4=5x-36-2\sqrt{4x-27}\sqrt{x-9}$$

Tenglamaning ikkala tarafidan 5x-36 ni ayirish.

$$x-4-\left(5x-36\right)=-2\sqrt{4x-27}\sqrt{x-9}$$

5x-36 teskarisini topish uchun har birining teskarisini toping.

$$x-4-5x+36=-2\sqrt{4x-27}\sqrt{x-9}$$

-4x ni olish uchun x va -5x ni birlashtirish.

$$-4x-4+36=-2\sqrt{4x-27}\sqrt{x-9}$$

32 olish uchun -4 va 36'ni qo'shing.

$$-4x+32=-2\sqrt{4x-27}\sqrt{x-9}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(-4x+32\right)^{2}=\left(-2\sqrt{4x-27}\sqrt{x-9}\right)^{2}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(-4x+32\right)^{2} kengaytirilishi uchun ishlating.

$$16x^{2}-256x+1024=\left(-2\sqrt{4x-27}\sqrt{x-9}\right)^{2}$$

\left(-2\sqrt{4x-27}\sqrt{x-9}\right)^{2} ni kengaytirish.

$$16x^{2}-256x+1024=\left(-2\right)^{2}\left(\sqrt{4x-27}\right)^{2}\left(\sqrt{x-9}\right)^{2}$$

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

$$16x^{2}-256x+1024=4\left(\sqrt{4x-27}\right)^{2}\left(\sqrt{x-9}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{4x-27} ga hisoblang va 4x-27 ni qiymatni oling.

$$16x^{2}-256x+1024=4\left(4x-27\right)\left(\sqrt{x-9}\right)^{2}$$

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

$$16x^{2}-256x+1024=4\left(4x-27\right)\left(x-9\right)$$

4 ga 4x-27 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$16x^{2}-256x+1024=\left(16x-108\right)\left(x-9\right)$$

16x-108 ifodaning har bir elementini x-9 ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$16x^{2}-256x+1024=16x^{2}-144x-108x+972$$

-252x ni olish uchun -144x va -108x ni birlashtirish.

$$16x^{2}-256x+1024=16x^{2}-252x+972$$

Ikkala tarafdan 16x^{2} ni ayirish.

$$16x^{2}-256x+1024-16x^{2}=-252x+972$$

0 ni olish uchun 16x^{2} va -16x^{2} ni birlashtirish.

$$-256x+1024=-252x+972$$

252x ni ikki tarafga qo’shing.

$$-256x+1024+252x=972$$

-4x ni olish uchun -256x va 252x ni birlashtirish.

$$-4x+1024=972$$

Ikkala tarafdan 1024 ni ayirish.

$$-4x=972-1024$$

-52 olish uchun 972 dan 1024 ni ayirish.

$$-4x=-52$$

Ikki tarafini -4 ga bo‘ling.

$$x=\frac{-52}{-4}$$

13 ni olish uchun -52 ni -4 ga bo‘ling.

$$x=13$$

\sqrt{x-4}-\sqrt{4x-27}+\sqrt{x-9}=0 tenglamasida x uchun 13 ni almashtiring.

$$\sqrt{13-4}-\sqrt{4\times 13-27}+\sqrt{13-9}=0$$

Qisqartirish. x=13 tenglamani qoniqtiradi.

$$0=0$$

\sqrt{x-4}=\sqrt{4x-27}-\sqrt{x-9} tenglamasi noyob yechimga ega.

$$x=13$$