sqrt{x+6}-sqrt{5x+1}+1=0 eching

$$\sqrt { x + 6 } - \sqrt { 5 x + 1 } + 1 = 0$$

x uchun yechish

$$x=3$$

Yechim qadamlari

Tenglamaning ikkala tarafidan -\sqrt{5x+1}+1 ni ayirish.

$$\sqrt{x+6}=-\left(-\sqrt{5x+1}+1\right)$$

-\sqrt{5x+1}+1 teskarisini topish uchun har birining teskarisini toping.

$$\sqrt{x+6}=-\left(-\sqrt{5x+1}\right)-1$$

-\sqrt{5x+1} ning teskarisi \sqrt{5x+1} ga teng.

$$\sqrt{x+6}=\sqrt{5x+1}-1$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{x+6}\right)^{2}=\left(\sqrt{5x+1}-1\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{x+6} ga hisoblang va x+6 ni qiymatni oling.

$$x+6=\left(\sqrt{5x+1}-1\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(\sqrt{5x+1}-1\right)^{2} kengaytirilishi uchun ishlating.

$$x+6=\left(\sqrt{5x+1}\right)^{2}-2\sqrt{5x+1}+1$$

2 daraja ko‘rsatkichini \sqrt{5x+1} ga hisoblang va 5x+1 ni qiymatni oling.

$$x+6=5x+1-2\sqrt{5x+1}+1$$

2 olish uchun 1 va 1'ni qo'shing.

$$x+6=5x+2-2\sqrt{5x+1}$$

Tenglamaning ikkala tarafidan 5x+2 ni ayirish.

$$x+6-\left(5x+2\right)=-2\sqrt{5x+1}$$

5x+2 teskarisini topish uchun har birining teskarisini toping.

$$x+6-5x-2=-2\sqrt{5x+1}$$

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

$$-4x+6-2=-2\sqrt{5x+1}$$

4 olish uchun 6 dan 2 ni ayirish.

$$-4x+4=-2\sqrt{5x+1}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(-4x+4\right)^{2}=\left(-2\sqrt{5x+1}\right)^{2}$$

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

$$16x^{2}-32x+16=\left(-2\sqrt{5x+1}\right)^{2}$$

\left(-2\sqrt{5x+1}\right)^{2} ni kengaytirish.

$$16x^{2}-32x+16=\left(-2\right)^{2}\left(\sqrt{5x+1}\right)^{2}$$

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

$$16x^{2}-32x+16=4\left(\sqrt{5x+1}\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{5x+1} ga hisoblang va 5x+1 ni qiymatni oling.

$$16x^{2}-32x+16=4\left(5x+1\right)$$

4 ga 5x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$16x^{2}-32x+16=20x+4$$

Ikkala tarafdan 20x ni ayirish.

$$16x^{2}-32x+16-20x=4$$

-52x ni olish uchun -32x va -20x ni birlashtirish.

$$16x^{2}-52x+16=4$$

Ikkala tarafdan 4 ni ayirish.

$$16x^{2}-52x+16-4=0$$

12 olish uchun 16 dan 4 ni ayirish.

$$16x^{2}-52x+12=0$$

Ikki tarafini 4 ga bo‘ling.

$$4x^{2}-13x+3=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 4x^{2}+ax+bx+3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-13 ab=4\times 3=12$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-12 -2,-6 -3,-4$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Yechim – -13 yigʻindisini beruvchi juftlik.

$$a=-12 b=-1$$

4x^{2}-13x+3 ni \left(4x^{2}-12x\right)+\left(-x+3\right) sifatida qaytadan yozish.

$$\left(4x^{2}-12x\right)+\left(-x+3\right)$$

Birinchi guruhda 4x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$4x\left(x-3\right)-\left(x-3\right)$$

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(4x-1\right)$$

Tenglamani yechish uchun x-3=0 va 4x-1=0 ni yeching.

$$x=3 x=\frac{1}{4}$$

\sqrt{x+6}-\sqrt{5x+1}+1=0 tenglamasida x uchun 3 ni almashtiring.

$$\sqrt{3+6}-\sqrt{5\times 3+1}+1=0$$

Qisqartirish. x=3 tenglamani qoniqtiradi.

$$0=0$$

\sqrt{x+6}-\sqrt{5x+1}+1=0 tenglamasida x uchun \frac{1}{4} ni almashtiring.

$$\sqrt{\frac{1}{4}+6}-\sqrt{5\times \frac{1}{4}+1}+1=0$$

Qisqartirish. x=\frac{1}{4} qiymati bu tenglamani qoniqtirmaydi.

$$2=0$$

\sqrt{x+6}-\sqrt{5x+1}+1=0 tenglamasida x uchun 3 ni almashtiring.

$$\sqrt{3+6}-\sqrt{5\times 3+1}+1=0$$

Qisqartirish. x=3 tenglamani qoniqtiradi.

$$0=0$$

\sqrt{x+6}=\sqrt{5x+1}-1 tenglamasi noyob yechimga ega.

$$x=3$$