10/(x-5)(x+1)+x/x+1=3/x-5 eching

$$\frac { 10 } { ( x - 5 ) ( x + 1 ) } + \frac { x } { x + 1 } = \frac { 3 } { x - 5 }$$

x uchun yechish

$$x=1$$

$$x=7$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -1,5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(x+1\right) ga, \left(x-5\right)\left(x+1\right),x+1,x-5 ning eng kichik karralisiga ko‘paytiring.

$$10+\left(x-5\right)x=\left(x+1\right)\times 3$$

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

$$10+x^{2}-5x=\left(x+1\right)\times 3$$

x+1 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10+x^{2}-5x=3x+3$$

Ikkala tarafdan 3x ni ayirish.

$$10+x^{2}-5x-3x=3$$

-8x ni olish uchun -5x va -3x ni birlashtirish.

$$10+x^{2}-8x=3$$

Ikkala tarafdan 3 ni ayirish.

$$10+x^{2}-8x-3=0$$

7 olish uchun 10 dan 3 ni ayirish.

$$7+x^{2}-8x=0$$

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.

$$x^{2}-8x+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -8 ni b va 7 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 7}}{2}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 7}}{2}$$

-4 ni 7 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-28}}{2}$$

64 ni -28 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{36}}{2}$$

36 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±6}{2}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±6}{2}$$

x=\frac{8±6}{2} tenglamasini yeching, bunda ± musbat. 8 ni 6 ga qo'shish.

$$x=\frac{14}{2}$$

14 ni 2 ga bo'lish.

$$x=7$$

x=\frac{8±6}{2} tenglamasini yeching, bunda ± manfiy. 8 dan 6 ni ayirish.

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

2 ni 2 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=7 x=1$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -1,5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(x+1\right) ga, \left(x-5\right)\left(x+1\right),x+1,x-5 ning eng kichik karralisiga ko‘paytiring.

$$10+\left(x-5\right)x=\left(x+1\right)\times 3$$

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

$$10+x^{2}-5x=\left(x+1\right)\times 3$$

x+1 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$10+x^{2}-5x=3x+3$$

Ikkala tarafdan 3x ni ayirish.

$$10+x^{2}-5x-3x=3$$

-8x ni olish uchun -5x va -3x ni birlashtirish.

$$10+x^{2}-8x=3$$

Ikkala tarafdan 10 ni ayirish.

$$x^{2}-8x=3-10$$

-7 olish uchun 3 dan 10 ni ayirish.

$$x^{2}-8x=-7$$

-8 ni bo‘lish, x shartining koeffitsienti, 2 ga -4 olish uchun. Keyin, -4 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-8x+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

-4 kvadratini chiqarish.

$$x^{2}-8x+16=-7+16$$

-7 ni 16 ga qo'shish.

$$x^{2}-8x+16=9$$

x^{2}-8x+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.

$$\left(x-4\right)^{2}=9$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{9}$$

Qisqartirish.

$$x-4=3 x-4=-3$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$x=7 x=1$$