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

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

x uchun yechish

$$x = \frac{\sqrt{53} + 3}{2} \approx 5,140054945$$

$$x=\frac{3-\sqrt{53}}{2}\approx -2,140054945$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -2,-1,1 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-1\right)\left(x+1\right)\left(x+2\right) ga, x^{3}+2x^{2}-x-2,1-x,x+1 ning eng kichik karralisiga ko‘paytiring.

$$1-\left(-\left(1+x\right)\left(2+x\right)\times 2\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-2 hosil qilish uchun -1 va 2 ni ko'paytirish.

$$1-\left(-2\left(1+x\right)\left(2+x\right)\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-2 ga 1+x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$1-\left(-2-2x\right)\left(2+x\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-2-2x ga 2+x ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$1-\left(-4-6x-2x^{2}\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-4-6x-2x^{2} teskarisini topish uchun har birining teskarisini toping.

$$1+4+6x+2x^{2}=\left(x-1\right)\left(x+2\right)\times 3$$

5 olish uchun 1 va 4'ni qo'shing.

$$5+6x+2x^{2}=\left(x-1\right)\left(x+2\right)\times 3$$

x-1 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

x^{2}+x-2 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5+6x+2x^{2}=3x^{2}+3x-6$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$5+6x+2x^{2}-3x^{2}=3x-6$$

-x^{2} ni olish uchun 2x^{2} va -3x^{2} ni birlashtirish.

$$5+6x-x^{2}=3x-6$$

Ikkala tarafdan 3x ni ayirish.

$$5+6x-x^{2}-3x=-6$$

3x ni olish uchun 6x va -3x ni birlashtirish.

$$5+3x-x^{2}=-6$$

6 ni ikki tarafga qo’shing.

$$5+3x-x^{2}+6=0$$

11 olish uchun 5 va 6'ni qo'shing.

$$11+3x-x^{2}=0$$

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.

$$-x^{2}+3x+11=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, 3 ni b va 11 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\times 11}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+4\times 11}}{2\left(-1\right)}$$

4 ni 11 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+44}}{2\left(-1\right)}$$

9 ni 44 ga qo'shish.

$$x=\frac{-3±\sqrt{53}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{53}}{-2}$$

x=\frac{-3±\sqrt{53}}{-2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{53} ga qo'shish.

$$x=\frac{\sqrt{53}-3}{-2}$$

-3+\sqrt{53} ni -2 ga bo'lish.

$$x=\frac{3-\sqrt{53}}{2}$$

x=\frac{-3±\sqrt{53}}{-2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{53} ni ayirish.

$$x=\frac{-\sqrt{53}-3}{-2}$$

-3-\sqrt{53} ni -2 ga bo'lish.

$$x=\frac{\sqrt{53}+3}{2}$$

Tenglama yechildi.

$$x=\frac{3-\sqrt{53}}{2} x=\frac{\sqrt{53}+3}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -2,-1,1 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-1\right)\left(x+1\right)\left(x+2\right) ga, x^{3}+2x^{2}-x-2,1-x,x+1 ning eng kichik karralisiga ko‘paytiring.

$$1-\left(-\left(1+x\right)\left(2+x\right)\times 2\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-2 hosil qilish uchun -1 va 2 ni ko'paytirish.

$$1-\left(-2\left(1+x\right)\left(2+x\right)\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-2 ga 1+x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$1-\left(-2-2x\right)\left(2+x\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-2-2x ga 2+x ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$1-\left(-4-6x-2x^{2}\right)=\left(x-1\right)\left(x+2\right)\times 3$$

-4-6x-2x^{2} teskarisini topish uchun har birining teskarisini toping.

$$1+4+6x+2x^{2}=\left(x-1\right)\left(x+2\right)\times 3$$

5 olish uchun 1 va 4'ni qo'shing.

$$5+6x+2x^{2}=\left(x-1\right)\left(x+2\right)\times 3$$

x-1 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

x^{2}+x-2 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5+6x+2x^{2}=3x^{2}+3x-6$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$5+6x+2x^{2}-3x^{2}=3x-6$$

-x^{2} ni olish uchun 2x^{2} va -3x^{2} ni birlashtirish.

$$5+6x-x^{2}=3x-6$$

Ikkala tarafdan 3x ni ayirish.

$$5+6x-x^{2}-3x=-6$$

3x ni olish uchun 6x va -3x ni birlashtirish.

$$5+3x-x^{2}=-6$$

Ikkala tarafdan 5 ni ayirish.

$$3x-x^{2}=-6-5$$

-11 olish uchun -6 dan 5 ni ayirish.

$$3x-x^{2}=-11$$

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

$$-x^{2}+3x=-11$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}+3x}{-1}=-\frac{11}{-1}$$

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

$$x^{2}+\frac{3}{-1}x=-\frac{11}{-1}$$

3 ni -1 ga bo'lish.

$$x^{2}-3x=-\frac{11}{-1}$$

-11 ni -1 ga bo'lish.

$$x^{2}-3x=11$$

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=11+\left(-\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$x^{2}-3x+\frac{9}{4}=11+\frac{9}{4}$$

11 ni \frac{9}{4} ga qo'shish.

$$x^{2}-3x+\frac{9}{4}=\frac{53}{4}$$

x^{2}-3x+\frac{9}{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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{53}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{53}}{2} x-\frac{3}{2}=-\frac{\sqrt{53}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{53}+3}{2} x=\frac{3-\sqrt{53}}{2}$$