b-2/b-1-5/b^2-4b+3+1=10/3-b eching

$$\frac { b - 2 } { b - 1 } - \frac { 5 } { b ^ { 2 } - 4 b + 3 } + 1 = \frac { 10 } { 3 - b }$$

b uchun yechish

$$b=-2$$

$$b = \frac{3}{2} = 1\frac{1}{2} = 1,5$$

Guruhlash orqali faktorlash qadamlari

b qiymati 1,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(b-3\right)\left(b-1\right) ga, b-1,b^{2}-4b+3,3-b ning eng kichik karralisiga ko‘paytiring.

$$\left(b-3\right)\left(b-2\right)-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

b-3 ga b-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$b^{2}-5b+6-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

1 olish uchun 6 dan 5 ni ayirish.

$$b^{2}-5b+1+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

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

$$b^{2}-5b+1+b^{2}-4b+3=\left(1-b\right)\times 10$$

2b^{2} ni olish uchun b^{2} va b^{2} ni birlashtirish.

$$2b^{2}-5b+1-4b+3=\left(1-b\right)\times 10$$

-9b ni olish uchun -5b va -4b ni birlashtirish.

$$2b^{2}-9b+1+3=\left(1-b\right)\times 10$$

4 olish uchun 1 va 3'ni qo'shing.

$$2b^{2}-9b+4=\left(1-b\right)\times 10$$

1-b ga 10 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2b^{2}-9b+4=10-10b$$

Ikkala tarafdan 10 ni ayirish.

$$2b^{2}-9b+4-10=-10b$$

-6 olish uchun 4 dan 10 ni ayirish.

$$2b^{2}-9b-6=-10b$$

10b ni ikki tarafga qo’shing.

$$2b^{2}-9b-6+10b=0$$

b ni olish uchun -9b va 10b ni birlashtirish.

$$2b^{2}+b-6=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 2b^{2}+ab+bb-6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=2\left(-6\right)=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

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

Har bir juftlik yigʻindisini hisoblang.

$$-1+12=11 -2+6=4 -3+4=1$$

Yechim – 1 yigʻindisini beruvchi juftlik.

$$a=-3 b=4$$

2b^{2}+b-6 ni \left(2b^{2}-3b\right)+\left(4b-6\right) sifatida qaytadan yozish.

$$\left(2b^{2}-3b\right)+\left(4b-6\right)$$

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

$$b\left(2b-3\right)+2\left(2b-3\right)$$

Distributiv funktsiyasidan foydalangan holda 2b-3 umumiy terminini chiqaring.

$$\left(2b-3\right)\left(b+2\right)$$

Tenglamani yechish uchun 2b-3=0 va b+2=0 ni yeching.

$$b=\frac{3}{2} b=-2$$

Kvadrat tenglamalaridan foydalanish qoidalari

b qiymati 1,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(b-3\right)\left(b-1\right) ga, b-1,b^{2}-4b+3,3-b ning eng kichik karralisiga ko‘paytiring.

$$\left(b-3\right)\left(b-2\right)-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

b-3 ga b-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$b^{2}-5b+6-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

1 olish uchun 6 dan 5 ni ayirish.

$$b^{2}-5b+1+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

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

$$b^{2}-5b+1+b^{2}-4b+3=\left(1-b\right)\times 10$$

2b^{2} ni olish uchun b^{2} va b^{2} ni birlashtirish.

$$2b^{2}-5b+1-4b+3=\left(1-b\right)\times 10$$

-9b ni olish uchun -5b va -4b ni birlashtirish.

$$2b^{2}-9b+1+3=\left(1-b\right)\times 10$$

4 olish uchun 1 va 3'ni qo'shing.

$$2b^{2}-9b+4=\left(1-b\right)\times 10$$

1-b ga 10 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2b^{2}-9b+4=10-10b$$

Ikkala tarafdan 10 ni ayirish.

$$2b^{2}-9b+4-10=-10b$$

-6 olish uchun 4 dan 10 ni ayirish.

$$2b^{2}-9b-6=-10b$$

10b ni ikki tarafga qo’shing.

$$2b^{2}-9b-6+10b=0$$

b ni olish uchun -9b va 10b ni birlashtirish.

$$2b^{2}+b-6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 1 ni b va -6 ni c bilan almashtiring.

$$b=\frac{-1±\sqrt{1^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

1 kvadratini chiqarish.

$$b=\frac{-1±\sqrt{1-4\times 2\left(-6\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$b=\frac{-1±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

-8 ni -6 marotabaga ko'paytirish.

$$b=\frac{-1±\sqrt{1+48}}{2\times 2}$$

1 ni 48 ga qo'shish.

$$b=\frac{-1±\sqrt{49}}{2\times 2}$$

49 ning kvadrat ildizini chiqarish.

$$b=\frac{-1±7}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$b=\frac{-1±7}{4}$$

b=\frac{-1±7}{4} tenglamasini yeching, bunda ± musbat. -1 ni 7 ga qo'shish.

$$b=\frac{6}{4}$$

\frac{6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$b=\frac{3}{2}$$

b=\frac{-1±7}{4} tenglamasini yeching, bunda ± manfiy. -1 dan 7 ni ayirish.

$$b=-\frac{8}{4}$$

-8 ni 4 ga bo'lish.

$$b=-2$$

Tenglama yechildi.

$$b=\frac{3}{2} b=-2$$

Kvadratni yakunlash bo'yicha qadamlar

b qiymati 1,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(b-3\right)\left(b-1\right) ga, b-1,b^{2}-4b+3,3-b ning eng kichik karralisiga ko‘paytiring.

$$\left(b-3\right)\left(b-2\right)-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

b-3 ga b-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$b^{2}-5b+6-5+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

1 olish uchun 6 dan 5 ni ayirish.

$$b^{2}-5b+1+\left(b-3\right)\left(b-1\right)=\left(1-b\right)\times 10$$

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

$$b^{2}-5b+1+b^{2}-4b+3=\left(1-b\right)\times 10$$

2b^{2} ni olish uchun b^{2} va b^{2} ni birlashtirish.

$$2b^{2}-5b+1-4b+3=\left(1-b\right)\times 10$$

-9b ni olish uchun -5b va -4b ni birlashtirish.

$$2b^{2}-9b+1+3=\left(1-b\right)\times 10$$

4 olish uchun 1 va 3'ni qo'shing.

$$2b^{2}-9b+4=\left(1-b\right)\times 10$$

1-b ga 10 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2b^{2}-9b+4=10-10b$$

10b ni ikki tarafga qo’shing.

$$2b^{2}-9b+4+10b=10$$

b ni olish uchun -9b va 10b ni birlashtirish.

$$2b^{2}+b+4=10$$

Ikkala tarafdan 4 ni ayirish.

$$2b^{2}+b=10-4$$

6 olish uchun 10 dan 4 ni ayirish.

$$2b^{2}+b=6$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2b^{2}+b}{2}=\frac{6}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$b^{2}+\frac{1}{2}b=\frac{6}{2}$$

6 ni 2 ga bo'lish.

$$b^{2}+\frac{1}{2}b=3$$

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$b^{2}+\frac{1}{2}b+\left(\frac{1}{4}\right)^{2}=3+\left(\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$b^{2}+\frac{1}{2}b+\frac{1}{16}=3+\frac{1}{16}$$

3 ni \frac{1}{16} ga qo'shish.

$$b^{2}+\frac{1}{2}b+\frac{1}{16}=\frac{49}{16}$$

b^{2}+\frac{1}{2}b+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(b+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(b+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Qisqartirish.

$$b+\frac{1}{4}=\frac{7}{4} b+\frac{1}{4}=-\frac{7}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$b=\frac{3}{2} b=-2$$