6v^2+17v+5 eching

$$6 v ^ { 2 } + 17 v + 5$$

Omil

$$\left(2v+5\right)\left(3v+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6v^{2}+av+bv+5 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=17 ab=6\times 5=30$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,30 2,15 3,10 5,6$$

Har bir juftlik yigʻindisini hisoblang.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

Yechim – 17 yigʻindisini beruvchi juftlik.

$$a=2 b=15$$

6v^{2}+17v+5 ni \left(6v^{2}+2v\right)+\left(15v+5\right) sifatida qaytadan yozish.

$$\left(6v^{2}+2v\right)+\left(15v+5\right)$$

Birinchi guruhda 2v ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$2v\left(3v+1\right)+5\left(3v+1\right)$$

Distributiv funktsiyasidan foydalangan holda 3v+1 umumiy terminini chiqaring.

$$\left(3v+1\right)\left(2v+5\right)$$

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.

$$6v^{2}+17v+5=0$$

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.

$$v=\frac{-17±\sqrt{17^{2}-4\times 6\times 5}}{2\times 6}$$

17 kvadratini chiqarish.

$$v=\frac{-17±\sqrt{289-4\times 6\times 5}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$v=\frac{-17±\sqrt{289-24\times 5}}{2\times 6}$$

-24 ni 5 marotabaga ko'paytirish.

$$v=\frac{-17±\sqrt{289-120}}{2\times 6}$$

289 ni -120 ga qo'shish.

$$v=\frac{-17±\sqrt{169}}{2\times 6}$$

169 ning kvadrat ildizini chiqarish.

$$v=\frac{-17±13}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$v=\frac{-17±13}{12}$$

v=\frac{-17±13}{12} tenglamasini yeching, bunda ± musbat. -17 ni 13 ga qo'shish.

$$v=-\frac{4}{12}$$

\frac{-4}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$v=-\frac{1}{3}$$

v=\frac{-17±13}{12} tenglamasini yeching, bunda ± manfiy. -17 dan 13 ni ayirish.

$$v=-\frac{30}{12}$$

\frac{-30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$v=-\frac{5}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{1}{3} ga va x_{2} uchun -\frac{5}{2} ga bo‘ling.

$$6v^{2}+17v+5=6\left(v-\left(-\frac{1}{3}\right)\right)\left(v-\left(-\frac{5}{2}\right)\right)$$

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

$$6v^{2}+17v+5=6\left(v+\frac{1}{3}\right)\left(v+\frac{5}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{3} ni v ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6v^{2}+17v+5=6\times \frac{3v+1}{3}\left(v+\frac{5}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni v ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6v^{2}+17v+5=6\times \frac{3v+1}{3}\times \frac{2v+5}{2}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3v+1}{3} ni \frac{2v+5}{2} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$6v^{2}+17v+5=6\times \frac{\left(3v+1\right)\left(2v+5\right)}{3\times 2}$$

3 ni 2 marotabaga ko'paytirish.

$$6v^{2}+17v+5=6\times \frac{\left(3v+1\right)\left(2v+5\right)}{6}$$

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

$$6v^{2}+17v+5=\left(3v+1\right)\left(2v+5\right)$$

Baholash

$$\left(2v+5\right)\left(3v+1\right)$$