1/3+1/x=1/2+x-1/6x eching

$$\frac { 1 } { 3 } + \frac { 1 } { x } = \frac { 1 } { 2 + x } - \frac { 1 } { 6 x }$$

x uchun yechish (complex solution)

$$x=\frac{-5+\sqrt{87}i}{4}\approx -1,25+2,331844763i$$

$$x=\frac{-\sqrt{87}i-5}{4}\approx -1,25-2,331844763i$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -2,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 6x\left(x+2\right) ga, 3,x,2+x,6x ning eng kichik karralisiga ko‘paytiring.

$$6x\left(x+2\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

6x ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(6x^{2}+12x\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

6x^{2}+12x ga \frac{1}{3} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}+4x+6x+12=6x-\left(x+2\right)$$

10x ni olish uchun 4x va 6x ni birlashtirish.

$$2x^{2}+10x+12=6x-\left(x+2\right)$$

x+2 teskarisini topish uchun har birining teskarisini toping.

$$2x^{2}+10x+12=6x-x-2$$

5x ni olish uchun 6x va -x ni birlashtirish.

$$2x^{2}+10x+12=5x-2$$

Ikkala tarafdan 5x ni ayirish.

$$2x^{2}+10x+12-5x=-2$$

5x ni olish uchun 10x va -5x ni birlashtirish.

$$2x^{2}+5x+12=-2$$

2 ni ikki tarafga qo’shing.

$$2x^{2}+5x+12+2=0$$

14 olish uchun 12 va 2'ni qo'shing.

$$2x^{2}+5x+14=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 5 ni b va 14 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\times 14}}{2\times 2}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 2\times 14}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-8\times 14}}{2\times 2}$$

-8 ni 14 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-112}}{2\times 2}$$

25 ni -112 ga qo'shish.

$$x=\frac{-5±\sqrt{-87}}{2\times 2}$$

-87 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±\sqrt{87}i}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{87}i}{4}$$

x=\frac{-5±\sqrt{87}i}{4} tenglamasini yeching, bunda ± musbat. -5 ni i\sqrt{87} ga qo'shish.

$$x=\frac{-5+\sqrt{87}i}{4}$$

x=\frac{-5±\sqrt{87}i}{4} tenglamasini yeching, bunda ± manfiy. -5 dan i\sqrt{87} ni ayirish.

$$x=\frac{-\sqrt{87}i-5}{4}$$

Tenglama yechildi.

$$x=\frac{-5+\sqrt{87}i}{4} x=\frac{-\sqrt{87}i-5}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -2,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 6x\left(x+2\right) ga, 3,x,2+x,6x ning eng kichik karralisiga ko‘paytiring.

$$6x\left(x+2\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

6x ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(6x^{2}+12x\right)\times \frac{1}{3}+6x+12=6x-\left(x+2\right)$$

6x^{2}+12x ga \frac{1}{3} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x^{2}+4x+6x+12=6x-\left(x+2\right)$$

10x ni olish uchun 4x va 6x ni birlashtirish.

$$2x^{2}+10x+12=6x-\left(x+2\right)$$

x+2 teskarisini topish uchun har birining teskarisini toping.

$$2x^{2}+10x+12=6x-x-2$$

5x ni olish uchun 6x va -x ni birlashtirish.

$$2x^{2}+10x+12=5x-2$$

Ikkala tarafdan 5x ni ayirish.

$$2x^{2}+10x+12-5x=-2$$

5x ni olish uchun 10x va -5x ni birlashtirish.

$$2x^{2}+5x+12=-2$$

Ikkala tarafdan 12 ni ayirish.

$$2x^{2}+5x=-2-12$$

-14 olish uchun -2 dan 12 ni ayirish.

$$2x^{2}+5x=-14$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+5x}{2}=-\frac{14}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{5}{2}x=-\frac{14}{2}$$

-14 ni 2 ga bo'lish.

$$x^{2}+\frac{5}{2}x=-7$$

\frac{5}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{4} olish uchun. Keyin, \frac{5}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=-7+\left(\frac{5}{4}\right)^{2}$$

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

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-7+\frac{25}{16}$$

-7 ni \frac{25}{16} ga qo'shish.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=-\frac{87}{16}$$

x^{2}+\frac{5}{2}x+\frac{25}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{4}\right)^{2}=-\frac{87}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{-\frac{87}{16}}$$

Qisqartirish.

$$x+\frac{5}{4}=\frac{\sqrt{87}i}{4} x+\frac{5}{4}=-\frac{\sqrt{87}i}{4}$$

Tenglamaning ikkala tarafidan \frac{5}{4} ni ayirish.

$$x=\frac{-5+\sqrt{87}i}{4} x=\frac{-\sqrt{87}i-5}{4}$$