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

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

x uchun yechish

$$x=38$$

$$x=2$$

Yechim qadamlari

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

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{2x+5} ga hisoblang va 2x+5 ni qiymatni oling.

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

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

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

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

$$2x+5=9+6\sqrt{x-2}+x-2$$

7 olish uchun 9 dan 2 ni ayirish.

$$2x+5=7+6\sqrt{x-2}+x$$

Tenglamaning ikkala tarafidan 7+x ni ayirish.

$$2x+5-\left(7+x\right)=6\sqrt{x-2}$$

7+x teskarisini topish uchun har birining teskarisini toping.

$$2x+5-7-x=6\sqrt{x-2}$$

-2 olish uchun 5 dan 7 ni ayirish.

$$2x-2-x=6\sqrt{x-2}$$

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

$$x-2=6\sqrt{x-2}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

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

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

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

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

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

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

$$x^{2}-4x+4=36\left(\sqrt{x-2}\right)^{2}$$

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

$$x^{2}-4x+4=36\left(x-2\right)$$

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

$$x^{2}-4x+4=36x-72$$

Ikkala tarafdan 36x ni ayirish.

$$x^{2}-4x+4-36x=-72$$

-40x ni olish uchun -4x va -36x ni birlashtirish.

$$x^{2}-40x+4=-72$$

72 ni ikki tarafga qo’shing.

$$x^{2}-40x+4+72=0$$

76 olish uchun 4 va 72'ni qo'shing.

$$x^{2}-40x+76=0$$

Bu tenglamani yechish uchun x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right) formulasi yordamida x^{2}-40x+76 ni faktorlang. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-40 ab=76$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 76-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-76 -2,-38 -4,-19$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-76=-77 -2-38=-40 -4-19=-23$$

Yechim – -40 yigʻindisini beruvchi juftlik.

$$a=-38 b=-2$$

Faktorlangan \left(x+a\right)\left(x+b\right) ifodani olingan qiymatlar bilan qaytadan yozing.

$$\left(x-38\right)\left(x-2\right)$$

Tenglamani yechish uchun x-38=0 va x-2=0 ni yeching.

$$x=38 x=2$$

\sqrt{2x+5}-\sqrt{x-2}=3 tenglamasida x uchun 38 ni almashtiring.

$$\sqrt{2\times 38+5}-\sqrt{38-2}=3$$

Qisqartirish. x=38 tenglamani qoniqtiradi.

$$3=3$$

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

$$\sqrt{2\times 2+5}-\sqrt{2-2}=3$$

Qisqartirish. x=2 tenglamani qoniqtiradi.

$$3=3$$

\sqrt{2x+5}=\sqrt{x-2}+3 boʻyicha barcha yechimlar roʻyxati.

$$x=38 x=2$$