p-1/p+3-2/p-3=7-3p/p^2-9 eching

$$\frac { p - 1 } { p + 3 } - \frac { 2 } { p - 3 } = \frac { 7 - 3 p } { p ^ { 2 } - 9 }$$

p uchun yechish

$$p=-2$$

$$p=5$$

Faktorlash qadamlari

p qiymati -3,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(p-3\right)\left(p+3\right) ga, p+3,p-3,p^{2}-9 ning eng kichik karralisiga ko‘paytiring.

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

p-3 ga p-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

p+3 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

2p+6 teskarisini topish uchun har birining teskarisini toping.

$$p^{2}-4p+3-2p-6=7-3p$$

-6p ni olish uchun -4p va -2p ni birlashtirish.

$$p^{2}-6p+3-6=7-3p$$

-3 olish uchun 3 dan 6 ni ayirish.

$$p^{2}-6p-3=7-3p$$

Ikkala tarafdan 7 ni ayirish.

$$p^{2}-6p-3-7=-3p$$

-10 olish uchun -3 dan 7 ni ayirish.

$$p^{2}-6p-10=-3p$$

3p ni ikki tarafga qo’shing.

$$p^{2}-6p-10+3p=0$$

-3p ni olish uchun -6p va 3p ni birlashtirish.

$$p^{2}-3p-10=0$$

Bu tenglamani yechish uchun p^{2}+\left(a+b\right)p+ab=\left(p+a\right)\left(p+b\right) formulasi yordamida p^{2}-3p-10 ni faktorlang. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=-10$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -10-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-10 2,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-10=-9 2-5=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=-5 b=2$$

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

$$\left(p-5\right)\left(p+2\right)$$

Tenglamani yechish uchun p-5=0 va p+2=0 ni yeching.

$$p=5 p=-2$$

Guruhlash orqali faktorlash qadamlari

p qiymati -3,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(p-3\right)\left(p+3\right) ga, p+3,p-3,p^{2}-9 ning eng kichik karralisiga ko‘paytiring.

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

p-3 ga p-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

p+3 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

2p+6 teskarisini topish uchun har birining teskarisini toping.

$$p^{2}-4p+3-2p-6=7-3p$$

-6p ni olish uchun -4p va -2p ni birlashtirish.

$$p^{2}-6p+3-6=7-3p$$

-3 olish uchun 3 dan 6 ni ayirish.

$$p^{2}-6p-3=7-3p$$

Ikkala tarafdan 7 ni ayirish.

$$p^{2}-6p-3-7=-3p$$

-10 olish uchun -3 dan 7 ni ayirish.

$$p^{2}-6p-10=-3p$$

3p ni ikki tarafga qo’shing.

$$p^{2}-6p-10+3p=0$$

-3p ni olish uchun -6p va 3p ni birlashtirish.

$$p^{2}-3p-10=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon p^{2}+ap+bp-10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=1\left(-10\right)=-10$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -10-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-10 2,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-10=-9 2-5=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=-5 b=2$$

p^{2}-3p-10 ni \left(p^{2}-5p\right)+\left(2p-10\right) sifatida qaytadan yozish.

$$\left(p^{2}-5p\right)+\left(2p-10\right)$$

Birinchi guruhda p ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$p\left(p-5\right)+2\left(p-5\right)$$

Distributiv funktsiyasidan foydalangan holda p-5 umumiy terminini chiqaring.

$$\left(p-5\right)\left(p+2\right)$$

Tenglamani yechish uchun p-5=0 va p+2=0 ni yeching.

$$p=5 p=-2$$

Kvadrat tenglamalaridan foydalanish qoidalari

p qiymati -3,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(p-3\right)\left(p+3\right) ga, p+3,p-3,p^{2}-9 ning eng kichik karralisiga ko‘paytiring.

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

p-3 ga p-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

p+3 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

2p+6 teskarisini topish uchun har birining teskarisini toping.

$$p^{2}-4p+3-2p-6=7-3p$$

-6p ni olish uchun -4p va -2p ni birlashtirish.

$$p^{2}-6p+3-6=7-3p$$

-3 olish uchun 3 dan 6 ni ayirish.

$$p^{2}-6p-3=7-3p$$

Ikkala tarafdan 7 ni ayirish.

$$p^{2}-6p-3-7=-3p$$

-10 olish uchun -3 dan 7 ni ayirish.

$$p^{2}-6p-10=-3p$$

3p ni ikki tarafga qo’shing.

$$p^{2}-6p-10+3p=0$$

-3p ni olish uchun -6p va 3p ni birlashtirish.

$$p^{2}-3p-10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -3 ni b va -10 ni c bilan almashtiring.

$$p=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)}}{2}$$

-3 kvadratini chiqarish.

$$p=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)}}{2}$$

-4 ni -10 marotabaga ko'paytirish.

$$p=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

9 ni 40 ga qo'shish.

$$p=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

49 ning kvadrat ildizini chiqarish.

$$p=\frac{-\left(-3\right)±7}{2}$$

-3 ning teskarisi 3 ga teng.

$$p=\frac{3±7}{2}$$

p=\frac{3±7}{2} tenglamasini yeching, bunda ± musbat. 3 ni 7 ga qo'shish.

$$p=\frac{10}{2}$$

10 ni 2 ga bo'lish.

$$p=5$$

p=\frac{3±7}{2} tenglamasini yeching, bunda ± manfiy. 3 dan 7 ni ayirish.

$$p=-\frac{4}{2}$$

-4 ni 2 ga bo'lish.

$$p=-2$$

Tenglama yechildi.

$$p=5 p=-2$$

Kvadratni yakunlash bo'yicha qadamlar

p qiymati -3,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(p-3\right)\left(p+3\right) ga, p+3,p-3,p^{2}-9 ning eng kichik karralisiga ko‘paytiring.

$$\left(p-3\right)\left(p-1\right)-\left(p+3\right)\times 2=7-3p$$

p-3 ga p-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$p^{2}-4p+3-\left(p+3\right)\times 2=7-3p$$

p+3 ga 2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$p^{2}-4p+3-\left(2p+6\right)=7-3p$$

2p+6 teskarisini topish uchun har birining teskarisini toping.

$$p^{2}-4p+3-2p-6=7-3p$$

-6p ni olish uchun -4p va -2p ni birlashtirish.

$$p^{2}-6p+3-6=7-3p$$

-3 olish uchun 3 dan 6 ni ayirish.

$$p^{2}-6p-3=7-3p$$

3p ni ikki tarafga qo’shing.

$$p^{2}-6p-3+3p=7$$

-3p ni olish uchun -6p va 3p ni birlashtirish.

$$p^{2}-3p-3=7$$

3 ni ikki tarafga qo’shing.

$$p^{2}-3p=7+3$$

10 olish uchun 7 va 3'ni qo'shing.

$$p^{2}-3p=10$$

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$p^{2}-3p+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$p^{2}-3p+\frac{9}{4}=10+\frac{9}{4}$$

10 ni \frac{9}{4} ga qo'shish.

$$p^{2}-3p+\frac{9}{4}=\frac{49}{4}$$

p^{2}-3p+\frac{9}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(p-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(p-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Qisqartirish.

$$p-\frac{3}{2}=\frac{7}{2} p-\frac{3}{2}=-\frac{7}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$p=5 p=-2$$