-1+6d^2-5d eching

$$- 1 + 6 d ^ { 2 } - 5 d$$

Omil

$$\left(d-1\right)\left(6d+1\right)$$

Yechim qadamlari

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$6d^{2}-5d-1$$

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6d^{2}+ad+bd-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=6\left(-1\right)=-6$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -6-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-6 2,-3$$

Har bir juftlik yigʻindisini hisoblang.

$$1-6=-5 2-3=-1$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=-6 b=1$$

6d^{2}-5d-1 ni \left(6d^{2}-6d\right)+\left(d-1\right) sifatida qaytadan yozish.

$$\left(6d^{2}-6d\right)+\left(d-1\right)$$

6d^{2}-6d ichida 6d ni ajrating.

$$6d\left(d-1\right)+d-1$$

Distributiv funktsiyasidan foydalangan holda d-1 umumiy terminini chiqaring.

$$\left(d-1\right)\left(6d+1\right)$$

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.

$$6d^{2}-5d-1=0$$

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.

$$d=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-1\right)}}{2\times 6}$$

-5 kvadratini chiqarish.

$$d=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-1\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$d=\frac{-\left(-5\right)±\sqrt{25-24\left(-1\right)}}{2\times 6}$$

-24 ni -1 marotabaga ko'paytirish.

$$d=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 6}$$

25 ni 24 ga qo'shish.

$$d=\frac{-\left(-5\right)±\sqrt{49}}{2\times 6}$$

49 ning kvadrat ildizini chiqarish.

$$d=\frac{-\left(-5\right)±7}{2\times 6}$$

-5 ning teskarisi 5 ga teng.

$$d=\frac{5±7}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$d=\frac{5±7}{12}$$

d=\frac{5±7}{12} tenglamasini yeching, bunda ± musbat. 5 ni 7 ga qo'shish.

$$d=\frac{12}{12}$$

12 ni 12 ga bo'lish.

$$d=1$$

d=\frac{5±7}{12} tenglamasini yeching, bunda ± manfiy. 5 dan 7 ni ayirish.

$$d=-\frac{2}{12}$$

\frac{-2}{12} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$d=-\frac{1}{6}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 1 ga va x_{2} uchun -\frac{1}{6} ga bo‘ling.

$$6d^{2}-5d-1=6\left(d-1\right)\left(d-\left(-\frac{1}{6}\right)\right)$$

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{6} ni d ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6d^{2}-5d-1=6\left(d-1\right)\times \frac{6d+1}{6}$$

6 va 6 ichida eng katta umumiy 6 faktorini bekor qiling.

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

Baholash

$$\left(d-1\right)\left(6d+1\right)$$