x^3-1/(x-1)^2-x^3+1/(x+1)^2=6 eching

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

x uchun yechish

$$x=2$$

$$x=-2$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

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

x^{2}+2x+1 ga x^{3}-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x-1\right)^{2}\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-1\right)^{2} kengaytirilishi uchun ishlating.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x^{2}-2x+1\right)\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

x^{2}-2x+1 ga x^{3}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x^{5}+x^{2}-2x^{4}-2x+x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

x^{5}+x^{2}-2x^{4}-2x+x^{3}+1 teskarisini topish uchun har birining teskarisini toping.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-x^{5}-x^{2}+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

0 ni olish uchun x^{5} va -x^{5} ni birlashtirish.

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

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

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

4x^{4} ni olish uchun 2x^{4} va 2x^{4} ni birlashtirish.

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

0 ni olish uchun -2x va 2x ni birlashtirish.

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

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

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

-2 olish uchun -1 dan 1 ni ayirish.

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-1\right)^{2} kengaytirilishi uchun ishlating.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+1\right)^{2} kengaytirilishi uchun ishlating.

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

6 ga x^{2}-2x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-2x^{2}+4x^{4}-2=\left(6x^{2}-12x+6\right)\left(x^{2}+2x+1\right)$$

6x^{2}-12x+6 ga x^{2}+2x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$-2x^{2}+4x^{4}-2=6x^{4}-12x^{2}+6$$

Ikkala tarafdan 6x^{4} ni ayirish.

$$-2x^{2}+4x^{4}-2-6x^{4}=-12x^{2}+6$$

-2x^{4} ni olish uchun 4x^{4} va -6x^{4} ni birlashtirish.

$$-2x^{2}-2x^{4}-2=-12x^{2}+6$$

12x^{2} ni ikki tarafga qo’shing.

$$-2x^{2}-2x^{4}-2+12x^{2}=6$$

10x^{2} ni olish uchun -2x^{2} va 12x^{2} ni birlashtirish.

$$10x^{2}-2x^{4}-2=6$$

Ikkala tarafdan 6 ni ayirish.

$$10x^{2}-2x^{4}-2-6=0$$

-8 olish uchun -2 dan 6 ni ayirish.

$$10x^{2}-2x^{4}-8=0$$

x^{2} uchun t ni almashtiring.

$$-2t^{2}+10t-8=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun -2 ni, b uchun 10 ni va c uchun -8 ni ayiring.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-2\right)\left(-8\right)}}{-2\times 2}$$

Hisoblarni amalga oshiring.

$$t=\frac{-10±6}{-4}$$

t=\frac{-10±6}{-4} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=1 t=4$$

x=t^{2} boʻlganda, yechimlar har bir t uchun x=±\sqrt{t} hisoblanishi orqali olinadi.

$$x=1 x=-1 x=2 x=-2$$

x qiymati 1,-1 qiymatlaridan birortasiga teng bo‘lmaydi.

$$x=-2 x=2$$