2*sqrt{x+1}+sqrt{x+2}=3 eching

$$2 \cdot \sqrt { x + 1 } + \sqrt { x + 2 } = 3$$

x uchun yechish

$$x=\frac{13-8\sqrt{3}}{3}\approx -0,28546882$$

Yechim qadamlari

Tenglamaning ikkala tarafidan \sqrt{x+2} ni ayirish.

$$2\sqrt{x+1}=3-\sqrt{x+2}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(2\sqrt{x+1}\right)^{2}=\left(3-\sqrt{x+2}\right)^{2}$$

\left(2\sqrt{x+1}\right)^{2} ni kengaytirish.

$$2^{2}\left(\sqrt{x+1}\right)^{2}=\left(3-\sqrt{x+2}\right)^{2}$$

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

$$4\left(\sqrt{x+1}\right)^{2}=\left(3-\sqrt{x+2}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x+1} ga hisoblang va x+1 ni qiymatni oling.

$$4\left(x+1\right)=\left(3-\sqrt{x+2}\right)^{2}$$

4 ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

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

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

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

$$4x+4=9-6\sqrt{x+2}+x+2$$

11 olish uchun 9 va 2'ni qo'shing.

$$4x+4=11-6\sqrt{x+2}+x$$

Tenglamaning ikkala tarafidan 11+x ni ayirish.

$$4x+4-\left(11+x\right)=-6\sqrt{x+2}$$

11+x teskarisini topish uchun har birining teskarisini toping.

$$4x+4-11-x=-6\sqrt{x+2}$$

-7 olish uchun 4 dan 11 ni ayirish.

$$4x-7-x=-6\sqrt{x+2}$$

3x ni olish uchun 4x va -x ni birlashtirish.

$$3x-7=-6\sqrt{x+2}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(3x-7\right)^{2}=\left(-6\sqrt{x+2}\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(3x-7\right)^{2} kengaytirilishi uchun ishlating.

$$9x^{2}-42x+49=\left(-6\sqrt{x+2}\right)^{2}$$

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

$$9x^{2}-42x+49=\left(-6\right)^{2}\left(\sqrt{x+2}\right)^{2}$$

2 daraja ko‘rsatkichini -6 ga hisoblang va 36 ni qiymatni oling.

$$9x^{2}-42x+49=36\left(\sqrt{x+2}\right)^{2}$$

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

$$9x^{2}-42x+49=36\left(x+2\right)$$

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

$$9x^{2}-42x+49=36x+72$$

Ikkala tarafdan 36x ni ayirish.

$$9x^{2}-42x+49-36x=72$$

-78x ni olish uchun -42x va -36x ni birlashtirish.

$$9x^{2}-78x+49=72$$

Ikkala tarafdan 72 ni ayirish.

$$9x^{2}-78x+49-72=0$$

-23 olish uchun 49 dan 72 ni ayirish.

$$9x^{2}-78x-23=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 9 ni a, -78 ni b va -23 ni c bilan almashtiring.

$$x=\frac{-\left(-78\right)±\sqrt{\left(-78\right)^{2}-4\times 9\left(-23\right)}}{2\times 9}$$

-78 kvadratini chiqarish.

$$x=\frac{-\left(-78\right)±\sqrt{6084-4\times 9\left(-23\right)}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-78\right)±\sqrt{6084-36\left(-23\right)}}{2\times 9}$$

-36 ni -23 marotabaga ko'paytirish.

$$x=\frac{-\left(-78\right)±\sqrt{6084+828}}{2\times 9}$$

6084 ni 828 ga qo'shish.

$$x=\frac{-\left(-78\right)±\sqrt{6912}}{2\times 9}$$

6912 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-78\right)±48\sqrt{3}}{2\times 9}$$

-78 ning teskarisi 78 ga teng.

$$x=\frac{78±48\sqrt{3}}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{78±48\sqrt{3}}{18}$$

x=\frac{78±48\sqrt{3}}{18} tenglamasini yeching, bunda ± musbat. 78 ni 48\sqrt{3} ga qo'shish.

$$x=\frac{48\sqrt{3}+78}{18}$$

78+48\sqrt{3} ni 18 ga bo'lish.

$$x=\frac{8\sqrt{3}+13}{3}$$

x=\frac{78±48\sqrt{3}}{18} tenglamasini yeching, bunda ± manfiy. 78 dan 48\sqrt{3} ni ayirish.

$$x=\frac{78-48\sqrt{3}}{18}$$

78-48\sqrt{3} ni 18 ga bo'lish.

$$x=\frac{13-8\sqrt{3}}{3}$$

Tenglama yechildi.

$$x=\frac{8\sqrt{3}+13}{3} x=\frac{13-8\sqrt{3}}{3}$$

2\sqrt{x+1}+\sqrt{x+2}=3 tenglamasida x uchun \frac{8\sqrt{3}+13}{3} ni almashtiring.

$$2\sqrt{\frac{8\sqrt{3}+13}{3}+1}+\sqrt{\frac{8\sqrt{3}+13}{3}+2}=3$$

Qisqartirish. x=\frac{8\sqrt{3}+13}{3} qiymati bu tenglamani qoniqtirmaydi.

$$5+\frac{8}{3}\times 3^{\frac{1}{2}}=3$$

2\sqrt{x+1}+\sqrt{x+2}=3 tenglamasida x uchun \frac{13-8\sqrt{3}}{3} ni almashtiring.

$$2\sqrt{\frac{13-8\sqrt{3}}{3}+1}+\sqrt{\frac{13-8\sqrt{3}}{3}+2}=3$$

Qisqartirish. x=\frac{13-8\sqrt{3}}{3} tenglamani qoniqtiradi.

$$3=3$$

2\sqrt{x+1}=-\sqrt{x+2}+3 tenglamasi noyob yechimga ega.

$$x=\frac{13-8\sqrt{3}}{3}$$