11f^2+13f+2 eching

$$11 f ^ { 2 } + 13 f + 2$$

Omil

$$\left(f+1\right)\left(11f+2\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 11f^{2}+af+bf+2 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=13 ab=11\times 2=22$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 22-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,22 2,11$$

Har bir juftlik yigʻindisini hisoblang.

$$1+22=23 2+11=13$$

Yechim – 13 yigʻindisini beruvchi juftlik.

$$a=2 b=11$$

11f^{2}+13f+2 ni \left(11f^{2}+2f\right)+\left(11f+2\right) sifatida qaytadan yozish.

$$\left(11f^{2}+2f\right)+\left(11f+2\right)$$

11f^{2}+2f ichida f ni ajrating.

$$f\left(11f+2\right)+11f+2$$

Distributiv funktsiyasidan foydalangan holda 11f+2 umumiy terminini chiqaring.

$$\left(11f+2\right)\left(f+1\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$11f^{2}+13f+2=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$f=\frac{-13±\sqrt{13^{2}-4\times 11\times 2}}{2\times 11}$$

13 kvadratini chiqarish.

$$f=\frac{-13±\sqrt{169-4\times 11\times 2}}{2\times 11}$$

-4 ni 11 marotabaga ko'paytirish.

$$f=\frac{-13±\sqrt{169-44\times 2}}{2\times 11}$$

-44 ni 2 marotabaga ko'paytirish.

$$f=\frac{-13±\sqrt{169-88}}{2\times 11}$$

169 ni -88 ga qo'shish.

$$f=\frac{-13±\sqrt{81}}{2\times 11}$$

81 ning kvadrat ildizini chiqarish.

$$f=\frac{-13±9}{2\times 11}$$

2 ni 11 marotabaga ko'paytirish.

$$f=\frac{-13±9}{22}$$

f=\frac{-13±9}{22} tenglamasini yeching, bunda ± musbat. -13 ni 9 ga qo'shish.

$$f=-\frac{4}{22}$$

\frac{-4}{22} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$f=-\frac{2}{11}$$

f=\frac{-13±9}{22} tenglamasini yeching, bunda ± manfiy. -13 dan 9 ni ayirish.

$$f=-\frac{22}{22}$$

-22 ni 22 ga bo'lish.

$$f=-1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{2}{11} ga va x_{2} uchun -1 ga bo‘ling.

$$11f^{2}+13f+2=11\left(f-\left(-\frac{2}{11}\right)\right)\left(f-\left(-1\right)\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$11f^{2}+13f+2=11\left(f+\frac{2}{11}\right)\left(f+1\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{2}{11} ni f ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$11f^{2}+13f+2=11\times \frac{11f+2}{11}\left(f+1\right)$$

11 va 11 ichida eng katta umumiy 11 faktorini bekor qiling.

$$11f^{2}+13f+2=\left(11f+2\right)\left(f+1\right)$$

Baholash

$$\left(f+1\right)\left(11f+2\right)$$