sqrt{2x-9}=5-sqrt{x-3} eching

$$\sqrt { 2 x - 9 } = 5 - \sqrt { x - 3 }$$

x uchun yechish

$$x=81-10\sqrt{53}\approx 8,198901107$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{2x-9}\right)^{2}=\left(5-\sqrt{x-3}\right)^{2}$$

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

$$2x-9=\left(5-\sqrt{x-3}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(5-\sqrt{x-3}\right)^{2} kengaytirilishi uchun ishlating.

$$2x-9=25-10\sqrt{x-3}+\left(\sqrt{x-3}\right)^{2}$$

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

$$2x-9=25-10\sqrt{x-3}+x-3$$

22 olish uchun 25 dan 3 ni ayirish.

$$2x-9=22-10\sqrt{x-3}+x$$

Tenglamaning ikkala tarafidan 22+x ni ayirish.

$$2x-9-\left(22+x\right)=-10\sqrt{x-3}$$

22+x teskarisini topish uchun har birining teskarisini toping.

$$2x-9-22-x=-10\sqrt{x-3}$$

-31 olish uchun -9 dan 22 ni ayirish.

$$2x-31-x=-10\sqrt{x-3}$$

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

$$x-31=-10\sqrt{x-3}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(x-31\right)^{2}=\left(-10\sqrt{x-3}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-31\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}-62x+961=\left(-10\sqrt{x-3}\right)^{2}$$

\left(-10\sqrt{x-3}\right)^{2} ni kengaytirish.

$$x^{2}-62x+961=\left(-10\right)^{2}\left(\sqrt{x-3}\right)^{2}$$

2 daraja ko‘rsatkichini -10 ga hisoblang va 100 ni qiymatni oling.

$$x^{2}-62x+961=100\left(\sqrt{x-3}\right)^{2}$$

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

$$x^{2}-62x+961=100\left(x-3\right)$$

100 ga x-3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{2}-62x+961=100x-300$$

Ikkala tarafdan 100x ni ayirish.

$$x^{2}-62x+961-100x=-300$$

-162x ni olish uchun -62x va -100x ni birlashtirish.

$$x^{2}-162x+961=-300$$

300 ni ikki tarafga qo’shing.

$$x^{2}-162x+961+300=0$$

1261 olish uchun 961 va 300'ni qo'shing.

$$x^{2}-162x+1261=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -162 ni b va 1261 ni c bilan almashtiring.

$$x=\frac{-\left(-162\right)±\sqrt{\left(-162\right)^{2}-4\times 1261}}{2}$$

-162 kvadratini chiqarish.

$$x=\frac{-\left(-162\right)±\sqrt{26244-4\times 1261}}{2}$$

-4 ni 1261 marotabaga ko'paytirish.

$$x=\frac{-\left(-162\right)±\sqrt{26244-5044}}{2}$$

26244 ni -5044 ga qo'shish.

$$x=\frac{-\left(-162\right)±\sqrt{21200}}{2}$$

21200 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-162\right)±20\sqrt{53}}{2}$$

-162 ning teskarisi 162 ga teng.

$$x=\frac{162±20\sqrt{53}}{2}$$

x=\frac{162±20\sqrt{53}}{2} tenglamasini yeching, bunda ± musbat. 162 ni 20\sqrt{53} ga qo'shish.

$$x=\frac{20\sqrt{53}+162}{2}$$

162+20\sqrt{53} ni 2 ga bo'lish.

$$x=10\sqrt{53}+81$$

x=\frac{162±20\sqrt{53}}{2} tenglamasini yeching, bunda ± manfiy. 162 dan 20\sqrt{53} ni ayirish.

$$x=\frac{162-20\sqrt{53}}{2}$$

162-20\sqrt{53} ni 2 ga bo'lish.

$$x=81-10\sqrt{53}$$

Tenglama yechildi.

$$x=10\sqrt{53}+81 x=81-10\sqrt{53}$$

\sqrt{2x-9}=5-\sqrt{x-3} tenglamasida x uchun 10\sqrt{53}+81 ni almashtiring.

$$\sqrt{2\left(10\sqrt{53}+81\right)-9}=5-\sqrt{10\sqrt{53}+81-3}$$

Qisqartirish. x=10\sqrt{53}+81 qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$10+53^{\frac{1}{2}}=-53^{\frac{1}{2}}$$

\sqrt{2x-9}=5-\sqrt{x-3} tenglamasida x uchun 81-10\sqrt{53} ni almashtiring.

$$\sqrt{2\left(81-10\sqrt{53}\right)-9}=5-\sqrt{81-10\sqrt{53}-3}$$

Qisqartirish. x=81-10\sqrt{53} tenglamani qoniqtiradi.

$$10-53^{\frac{1}{2}}=10-53^{\frac{1}{2}}$$

\sqrt{2x-9}=-\sqrt{x-3}+5 tenglamasi noyob yechimga ega.

$$x=81-10\sqrt{53}$$