5/x+1-1=10-x/x-1 eching

$$\frac { 5 } { x + 1 } - 1 = \frac { 10 - x } { x - 1 }$$

x uchun yechish

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3,5$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

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

$$\left(x-1\right)\times 5+\left(x-1\right)\left(x+1\right)\left(-1\right)=\left(x+1\right)\left(10-x\right)$$

x-1 ga 5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x-5+\left(x-1\right)\left(x+1\right)\left(-1\right)=\left(x+1\right)\left(10-x\right)$$

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

$$5x-5+\left(x^{2}-1\right)\left(-1\right)=\left(x+1\right)\left(10-x\right)$$

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

$$5x-5-x^{2}+1=\left(x+1\right)\left(10-x\right)$$

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

$$5x-4-x^{2}=\left(x+1\right)\left(10-x\right)$$

x+1 ga 10-x ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5x-4-x^{2}=9x-x^{2}+10$$

Ikkala tarafdan 9x ni ayirish.

$$5x-4-x^{2}-9x=-x^{2}+10$$

-4x ni olish uchun 5x va -9x ni birlashtirish.

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

x^{2} ni ikki tarafga qo’shing.

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

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

$$-4x-4=10$$

4 ni ikki tarafga qo’shing.

$$-4x=10+4$$

14 olish uchun 10 va 4'ni qo'shing.

$$-4x=14$$

Ikki tarafini -4 ga bo‘ling.

$$x=\frac{14}{-4}$$

\frac{14}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{7}{2}$$