-f/10f+42=1/f+3 eching

$$\frac { - f } { 10 f + 42 } = \frac { 1 } { f + 3 }$$

f uchun yechish

$$f=-7$$

$$f=-6$$

Kvadrat tenglamalaridan foydalanish qoidalari

f qiymati -\frac{21}{5},-3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2\left(f+3\right)\left(5f+21\right) ga, 10f+42,f+3 ning eng kichik karralisiga ko‘paytiring.

$$\left(f+3\right)\left(-f\right)=10f+42$$

f+3 ga -f ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$f\left(-f\right)+3\left(-f\right)=10f+42$$

Ikkala tarafdan 10f ni ayirish.

$$f\left(-f\right)+3\left(-f\right)-10f=42$$

Ikkala tarafdan 42 ni ayirish.

$$f\left(-f\right)+3\left(-f\right)-10f-42=0$$

f^{2} hosil qilish uchun f va f ni ko'paytirish.

$$f^{2}\left(-1\right)+3\left(-1\right)f-10f-42=0$$

-3 hosil qilish uchun 3 va -1 ni ko'paytirish.

$$f^{2}\left(-1\right)-3f-10f-42=0$$

-13f ni olish uchun -3f va -10f ni birlashtirish.

$$f^{2}\left(-1\right)-13f-42=0$$

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.

$$-f^{2}-13f-42=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -13 ni b va -42 ni c bilan almashtiring.

$$f=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-1\right)\left(-42\right)}}{2\left(-1\right)}$$

-13 kvadratini chiqarish.

$$f=\frac{-\left(-13\right)±\sqrt{169-4\left(-1\right)\left(-42\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$f=\frac{-\left(-13\right)±\sqrt{169+4\left(-42\right)}}{2\left(-1\right)}$$

4 ni -42 marotabaga ko'paytirish.

$$f=\frac{-\left(-13\right)±\sqrt{169-168}}{2\left(-1\right)}$$

169 ni -168 ga qo'shish.

$$f=\frac{-\left(-13\right)±\sqrt{1}}{2\left(-1\right)}$$

1 ning kvadrat ildizini chiqarish.

$$f=\frac{-\left(-13\right)±1}{2\left(-1\right)}$$

-13 ning teskarisi 13 ga teng.

$$f=\frac{13±1}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$f=\frac{13±1}{-2}$$

f=\frac{13±1}{-2} tenglamasini yeching, bunda ± musbat. 13 ni 1 ga qo'shish.

$$f=\frac{14}{-2}$$

14 ni -2 ga bo'lish.

$$f=-7$$

f=\frac{13±1}{-2} tenglamasini yeching, bunda ± manfiy. 13 dan 1 ni ayirish.

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

12 ni -2 ga bo'lish.

$$f=-6$$

Tenglama yechildi.

$$f=-7 f=-6$$

Kvadratni yakunlash bo'yicha qadamlar

f qiymati -\frac{21}{5},-3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2\left(f+3\right)\left(5f+21\right) ga, 10f+42,f+3 ning eng kichik karralisiga ko‘paytiring.

$$\left(f+3\right)\left(-f\right)=10f+42$$

f+3 ga -f ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$f\left(-f\right)+3\left(-f\right)=10f+42$$

Ikkala tarafdan 10f ni ayirish.

$$f\left(-f\right)+3\left(-f\right)-10f=42$$

f^{2} hosil qilish uchun f va f ni ko'paytirish.

$$f^{2}\left(-1\right)+3\left(-1\right)f-10f=42$$

-3 hosil qilish uchun 3 va -1 ni ko'paytirish.

$$f^{2}\left(-1\right)-3f-10f=42$$

-13f ni olish uchun -3f va -10f ni birlashtirish.

$$f^{2}\left(-1\right)-13f=42$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-f^{2}-13f=42$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-f^{2}-13f}{-1}=\frac{42}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$f^{2}+\left(-\frac{13}{-1}\right)f=\frac{42}{-1}$$

-13 ni -1 ga bo'lish.

$$f^{2}+13f=\frac{42}{-1}$$

42 ni -1 ga bo'lish.

$$f^{2}+13f=-42$$

13 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{13}{2} olish uchun. Keyin, \frac{13}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$f^{2}+13f+\left(\frac{13}{2}\right)^{2}=-42+\left(\frac{13}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{13}{2} kvadratini chiqarish.

$$f^{2}+13f+\frac{169}{4}=-42+\frac{169}{4}$$

-42 ni \frac{169}{4} ga qo'shish.

$$f^{2}+13f+\frac{169}{4}=\frac{1}{4}$$

f^{2}+13f+\frac{169}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(f+\frac{13}{2}\right)^{2}=\frac{1}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(f+\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Qisqartirish.

$$f+\frac{13}{2}=\frac{1}{2} f+\frac{13}{2}=-\frac{1}{2}$$

Tenglamaning ikkala tarafidan \frac{13}{2} ni ayirish.

$$f=-6 f=-7$$