x+6/x-5+x-5/x+6=2x^2+23x+4/x^2+x-30 eching

$$\frac { x + 6 } { x - 5 } + \frac { x - 5 } { x + 6 } = \frac { 2 x ^ { 2 } + 23 x + 4 } { x ^ { 2 } + x - 30 }$$

x uchun yechish

$$x = \frac{19}{7} = 2\frac{5}{7} \approx 2,714285714$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

x qiymati -6,5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(x+6\right) ga, x-5,x+6,x^{2}+x-30 ning eng kichik karralisiga ko‘paytiring.

$$\left(x+6\right)\left(x+6\right)+\left(x-5\right)\left(x-5\right)=2x^{2}+23x+4$$

\left(x+6\right)^{2} hosil qilish uchun x+6 va x+6 ni ko'paytirish.

$$\left(x+6\right)^{2}+\left(x-5\right)\left(x-5\right)=2x^{2}+23x+4$$

\left(x-5\right)^{2} hosil qilish uchun x-5 va x-5 ni ko'paytirish.

$$\left(x+6\right)^{2}+\left(x-5\right)^{2}=2x^{2}+23x+4$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+6\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}+12x+36+\left(x-5\right)^{2}=2x^{2}+23x+4$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}+12x+36+x^{2}-10x+25=2x^{2}+23x+4$$

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

$$2x^{2}+12x+36-10x+25=2x^{2}+23x+4$$

2x ni olish uchun 12x va -10x ni birlashtirish.

$$2x^{2}+2x+36+25=2x^{2}+23x+4$$

61 olish uchun 36 va 25'ni qo'shing.

$$2x^{2}+2x+61=2x^{2}+23x+4$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$2x^{2}+2x+61-2x^{2}=23x+4$$

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

$$2x+61=23x+4$$

Ikkala tarafdan 23x ni ayirish.

$$2x+61-23x=4$$

-21x ni olish uchun 2x va -23x ni birlashtirish.

$$-21x+61=4$$

Ikkala tarafdan 61 ni ayirish.

$$-21x=4-61$$

-57 olish uchun 4 dan 61 ni ayirish.

$$-21x=-57$$

Ikki tarafini -21 ga bo‘ling.

$$x=\frac{-57}{-21}$$

\frac{-57}{-21} ulushini -3 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{19}{7}$$