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

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

x uchun yechish

$$x=5$$

$$x=1$$

Yechim qadamlari

Tenglamaning ikkala tarafidan -\sqrt{2x-1} ni ayirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{3x+1} ga hisoblang va 3x+1 ni qiymatni oling.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(1+\sqrt{2x-1}\right)^{2} kengaytirilishi uchun ishlating.

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

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

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

0 olish uchun 1 dan 1 ni ayirish.

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

Tenglamaning ikkala tarafidan 2x ni ayirish.

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

x ni olish uchun 3x va -2x ni birlashtirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

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

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

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

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

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

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

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

4 ga 2x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{2}+2x+1=8x-4$$

Ikkala tarafdan 8x ni ayirish.

$$x^{2}+2x+1-8x=-4$$

-6x ni olish uchun 2x va -8x ni birlashtirish.

$$x^{2}-6x+1=-4$$

4 ni ikki tarafga qo’shing.

$$x^{2}-6x+1+4=0$$

5 olish uchun 1 va 4'ni qo'shing.

$$x^{2}-6x+5=0$$

Bu tenglamani yechish uchun x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right) formulasi yordamida x^{2}-6x+5 ni faktorlang. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-6 ab=5$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. Faqat bundan juftlik tizim yechimidir.

$$a=-5 b=-1$$

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

$$\left(x-5\right)\left(x-1\right)$$

Tenglamani yechish uchun x-5=0 va x-1=0 ni yeching.

$$x=5 x=1$$

\sqrt{3x+1}-\sqrt{2x-1}=1 tenglamasida x uchun 5 ni almashtiring.

$$\sqrt{3\times 5+1}-\sqrt{2\times 5-1}=1$$

Qisqartirish. x=5 tenglamani qoniqtiradi.

$$1=1$$

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

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

Qisqartirish. x=1 tenglamani qoniqtiradi.

$$1=1$$

\sqrt{3x+1}=\sqrt{2x-1}+1 boʻyicha barcha yechimlar roʻyxati.

$$x=5 x=1$$