1/3x^2+4/5x=1 eching

$$\frac { 1 } { 3 } x ^ { 2 } + \frac { 4 } { 5 } x = 1$$

x uchun yechish

$$x=\frac{\sqrt{111}-6}{5}\approx 0,907130751$$

$$x=\frac{-\sqrt{111}-6}{5}\approx -3,307130751$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$\frac{1}{3}x^{2}+\frac{4}{5}x=1$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$\frac{1}{3}x^{2}+\frac{4}{5}x-1=1-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$\frac{1}{3}x^{2}+\frac{4}{5}x-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} \frac{1}{3} ni a, \frac{4}{5} ni b va -1 ni c bilan almashtiring.

$$x=\frac{-\frac{4}{5}±\sqrt{\left(\frac{4}{5}\right)^{2}-4\times \frac{1}{3}\left(-1\right)}}{2\times \frac{1}{3}}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{4}{5} kvadratini chiqarish.

$$x=\frac{-\frac{4}{5}±\sqrt{\frac{16}{25}-4\times \frac{1}{3}\left(-1\right)}}{2\times \frac{1}{3}}$$

-4 ni \frac{1}{3} marotabaga ko'paytirish.

$$x=\frac{-\frac{4}{5}±\sqrt{\frac{16}{25}-\frac{4}{3}\left(-1\right)}}{2\times \frac{1}{3}}$$

-\frac{4}{3} ni -1 marotabaga ko'paytirish.

$$x=\frac{-\frac{4}{5}±\sqrt{\frac{16}{25}+\frac{4}{3}}}{2\times \frac{1}{3}}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{16}{25} ni \frac{4}{3} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x=\frac{-\frac{4}{5}±\sqrt{\frac{148}{75}}}{2\times \frac{1}{3}}$$

\frac{148}{75} ning kvadrat ildizini chiqarish.

$$x=\frac{-\frac{4}{5}±\frac{2\sqrt{111}}{15}}{2\times \frac{1}{3}}$$

2 ni \frac{1}{3} marotabaga ko'paytirish.

$$x=\frac{-\frac{4}{5}±\frac{2\sqrt{111}}{15}}{\frac{2}{3}}$$

x=\frac{-\frac{4}{5}±\frac{2\sqrt{111}}{15}}{\frac{2}{3}} tenglamasini yeching, bunda ± musbat. -\frac{4}{5} ni \frac{2\sqrt{111}}{15} ga qo'shish.

$$x=\frac{\frac{2\sqrt{111}}{15}-\frac{4}{5}}{\frac{2}{3}}$$

-\frac{4}{5}+\frac{2\sqrt{111}}{15} ni \frac{2}{3} ga bo'lish -\frac{4}{5}+\frac{2\sqrt{111}}{15} ga k'paytirish \frac{2}{3} ga qaytarish.

$$x=\frac{\sqrt{111}-6}{5}$$

x=\frac{-\frac{4}{5}±\frac{2\sqrt{111}}{15}}{\frac{2}{3}} tenglamasini yeching, bunda ± manfiy. -\frac{4}{5} dan \frac{2\sqrt{111}}{15} ni ayirish.

$$x=\frac{-\frac{2\sqrt{111}}{15}-\frac{4}{5}}{\frac{2}{3}}$$

-\frac{4}{5}-\frac{2\sqrt{111}}{15} ni \frac{2}{3} ga bo'lish -\frac{4}{5}-\frac{2\sqrt{111}}{15} ga k'paytirish \frac{2}{3} ga qaytarish.

$$x=\frac{-\sqrt{111}-6}{5}$$

Tenglama yechildi.

$$x=\frac{\sqrt{111}-6}{5} x=\frac{-\sqrt{111}-6}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$\frac{1}{3}x^{2}+\frac{4}{5}x=1$$

Ikkala tarafini 3 ga ko‘paytiring.

$$\frac{\frac{1}{3}x^{2}+\frac{4}{5}x}{\frac{1}{3}}=\frac{1}{\frac{1}{3}}$$

\frac{1}{3} ga bo'lish \frac{1}{3} ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{\frac{4}{5}}{\frac{1}{3}}x=\frac{1}{\frac{1}{3}}$$

\frac{4}{5} ni \frac{1}{3} ga bo'lish \frac{4}{5} ga k'paytirish \frac{1}{3} ga qaytarish.

$$x^{2}+\frac{12}{5}x=\frac{1}{\frac{1}{3}}$$

1 ni \frac{1}{3} ga bo'lish 1 ga k'paytirish \frac{1}{3} ga qaytarish.

$$x^{2}+\frac{12}{5}x=3$$

\frac{12}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{6}{5} olish uchun. Keyin, \frac{6}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=3+\left(\frac{6}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{6}{5} kvadratini chiqarish.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=3+\frac{36}{25}$$

3 ni \frac{36}{25} ga qo'shish.

$$x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{111}{25}$$

x^{2}+\frac{12}{5}x+\frac{36}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{6}{5}\right)^{2}=\frac{111}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{111}{25}}$$

Qisqartirish.

$$x+\frac{6}{5}=\frac{\sqrt{111}}{5} x+\frac{6}{5}=-\frac{\sqrt{111}}{5}$$

Tenglamaning ikkala tarafidan \frac{6}{5} ni ayirish.

$$x=\frac{\sqrt{111}-6}{5} x=\frac{-\sqrt{111}-6}{5}$$