2f^2+5f eching

$$2 f ^ { 2 } + 5 f$$

Omil

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

Yechim qadamlari

f omili.

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

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.

$$2f^{2}+5f=0$$

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.

$$f=\frac{-5±\sqrt{5^{2}}}{2\times 2}$$

5^{2} ning kvadrat ildizini chiqarish.

$$f=\frac{-5±5}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$f=\frac{-5±5}{4}$$

f=\frac{-5±5}{4} tenglamasini yeching, bunda ± musbat. -5 ni 5 ga qo'shish.

$$f=\frac{0}{4}$$

0 ni 4 ga bo'lish.

$$f=0$$

f=\frac{-5±5}{4} tenglamasini yeching, bunda ± manfiy. -5 dan 5 ni ayirish.

$$f=-\frac{10}{4}$$

\frac{-10}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

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.

$$2f^{2}+5f=2f\left(f-\left(-\frac{5}{2}\right)\right)$$

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

$$2f^{2}+5f=2f\left(f+\frac{5}{2}\right)$$

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

$$2f^{2}+5f=2f\times \frac{2f+5}{2}$$

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

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

Baholash

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