6r^2-11r+4 eching

$$6 r ^ { 2 } - 11 r + 4$$

Omil

$$\left(2r-1\right)\left(3r-4\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6r^{2}+ar+br+4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=6\times 4=24$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 24-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-8 b=-3$$

6r^{2}-11r+4 ni \left(6r^{2}-8r\right)+\left(-3r+4\right) sifatida qaytadan yozish.

$$\left(6r^{2}-8r\right)+\left(-3r+4\right)$$

Birinchi guruhda 2r ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$2r\left(3r-4\right)-\left(3r-4\right)$$

Distributiv funktsiyasidan foydalangan holda 3r-4 umumiy terminini chiqaring.

$$\left(3r-4\right)\left(2r-1\right)$$

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.

$$6r^{2}-11r+4=0$$

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.

$$r=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\times 4}}{2\times 6}$$

-11 kvadratini chiqarish.

$$r=\frac{-\left(-11\right)±\sqrt{121-4\times 6\times 4}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$r=\frac{-\left(-11\right)±\sqrt{121-24\times 4}}{2\times 6}$$

-24 ni 4 marotabaga ko'paytirish.

$$r=\frac{-\left(-11\right)±\sqrt{121-96}}{2\times 6}$$

121 ni -96 ga qo'shish.

$$r=\frac{-\left(-11\right)±\sqrt{25}}{2\times 6}$$

25 ning kvadrat ildizini chiqarish.

$$r=\frac{-\left(-11\right)±5}{2\times 6}$$

-11 ning teskarisi 11 ga teng.

$$r=\frac{11±5}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$r=\frac{11±5}{12}$$

r=\frac{11±5}{12} tenglamasini yeching, bunda ± musbat. 11 ni 5 ga qo'shish.

$$r=\frac{16}{12}$$

\frac{16}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$r=\frac{4}{3}$$

r=\frac{11±5}{12} tenglamasini yeching, bunda ± manfiy. 11 dan 5 ni ayirish.

$$r=\frac{6}{12}$$

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

$$r=\frac{1}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{4}{3} ga va x_{2} uchun \frac{1}{2} ga bo‘ling.

$$6r^{2}-11r+4=6\left(r-\frac{4}{3}\right)\left(r-\frac{1}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{4}{3} ni r dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$6r^{2}-11r+4=6\times \frac{3r-4}{3}\left(r-\frac{1}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{1}{2} ni r dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$6r^{2}-11r+4=6\times \frac{3r-4}{3}\times \frac{2r-1}{2}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3r-4}{3} ni \frac{2r-1}{2} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$6r^{2}-11r+4=6\times \frac{\left(3r-4\right)\left(2r-1\right)}{3\times 2}$$

3 ni 2 marotabaga ko'paytirish.

$$6r^{2}-11r+4=6\times \frac{\left(3r-4\right)\left(2r-1\right)}{6}$$

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

$$6r^{2}-11r+4=\left(3r-4\right)\left(2r-1\right)$$

Baholash

$$\left(2r-1\right)\left(3r-4\right)$$