10c^2+25c eching

$$10 c ^ { 2 } + 25 c$$

Omil

$$5c\left(2c+5\right)$$

Yechim qadamlari

5 omili.

$$5\left(2c^{2}+5c\right)$$

Hisoblang: 2c^{2}+5c. c omili.

$$c\left(2c+5\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$5c\left(2c+5\right)$$

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.

$$10c^{2}+25c=0$$

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.

$$c=\frac{-25±\sqrt{25^{2}}}{2\times 10}$$

25^{2} ning kvadrat ildizini chiqarish.

$$c=\frac{-25±25}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$c=\frac{-25±25}{20}$$

c=\frac{-25±25}{20} tenglamasini yeching, bunda ± musbat. -25 ni 25 ga qo'shish.

$$c=\frac{0}{20}$$

0 ni 20 ga bo'lish.

$$c=0$$

c=\frac{-25±25}{20} tenglamasini yeching, bunda ± manfiy. -25 dan 25 ni ayirish.

$$c=-\frac{50}{20}$$

\frac{-50}{20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -\frac{5}{2} ga bo‘ling.

$$10c^{2}+25c=10c\left(c-\left(-\frac{5}{2}\right)\right)$$

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

$$10c^{2}+25c=10c\left(c+\frac{5}{2}\right)$$

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

$$10c^{2}+25c=10c\times \frac{2c+5}{2}$$

10 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$10c^{2}+25c=5c\left(2c+5\right)$$

Baholash

$$5c\left(2c+5\right)$$