x/x+1+x+1/x=13/6 eching

$$\frac { x } { x + 1 } + \frac { x + 1 } { x } = \frac { 13 } { 6 }$$

x uchun yechish

$$x=-3$$

$$x=2$$

Guruhlash orqali faktorlash qadamlari

x qiymati -1,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 6x\left(x+1\right) ga, x+1,x,6 ning eng kichik karralisiga ko‘paytiring.

$$6xx+\left(6x+6\right)\left(x+1\right)=13x\left(x+1\right)$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$6x^{2}+\left(6x+6\right)\left(x+1\right)=13x\left(x+1\right)$$

6x+6 ga x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$6x^{2}+6x^{2}+12x+6=13x\left(x+1\right)$$

12x^{2} ni olish uchun 6x^{2} va 6x^{2} ni birlashtirish.

$$12x^{2}+12x+6=13x\left(x+1\right)$$

13x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x^{2}+12x+6=13x^{2}+13x$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$12x^{2}+12x+6-13x^{2}=13x$$

-x^{2} ni olish uchun 12x^{2} va -13x^{2} ni birlashtirish.

$$-x^{2}+12x+6=13x$$

Ikkala tarafdan 13x ni ayirish.

$$-x^{2}+12x+6-13x=0$$

-x ni olish uchun 12x va -13x ni birlashtirish.

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

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -x^{2}+ax+bx+6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=-6=-6$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -6-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-6 2,-3$$

Har bir juftlik yigʻindisini hisoblang.

$$1-6=-5 2-3=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=2 b=-3$$

-x^{2}-x+6 ni \left(-x^{2}+2x\right)+\left(-3x+6\right) sifatida qaytadan yozish.

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

Birinchi guruhda x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$x\left(-x+2\right)+3\left(-x+2\right)$$

Distributiv funktsiyasidan foydalangan holda -x+2 umumiy terminini chiqaring.

$$\left(-x+2\right)\left(x+3\right)$$

Tenglamani yechish uchun -x+2=0 va x+3=0 ni yeching.

$$x=2 x=-3$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -1,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 6x\left(x+1\right) ga, x+1,x,6 ning eng kichik karralisiga ko‘paytiring.

$$6xx+\left(6x+6\right)\left(x+1\right)=13x\left(x+1\right)$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$6x^{2}+\left(6x+6\right)\left(x+1\right)=13x\left(x+1\right)$$

6x+6 ga x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$6x^{2}+6x^{2}+12x+6=13x\left(x+1\right)$$

12x^{2} ni olish uchun 6x^{2} va 6x^{2} ni birlashtirish.

$$12x^{2}+12x+6=13x\left(x+1\right)$$

13x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x^{2}+12x+6=13x^{2}+13x$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$12x^{2}+12x+6-13x^{2}=13x$$

-x^{2} ni olish uchun 12x^{2} va -13x^{2} ni birlashtirish.

$$-x^{2}+12x+6=13x$$

Ikkala tarafdan 13x ni ayirish.

$$-x^{2}+12x+6-13x=0$$

-x ni olish uchun 12x va -13x ni birlashtirish.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -1 ni b va 6 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 6}}{2\left(-1\right)}$$

4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-1\right)}$$

1 ni 24 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-1\right)}$$

25 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±5}{2\left(-1\right)}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±5}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{1±5}{-2}$$

x=\frac{1±5}{-2} tenglamasini yeching, bunda ± musbat. 1 ni 5 ga qo'shish.

$$x=\frac{6}{-2}$$

6 ni -2 ga bo'lish.

$$x=-3$$

x=\frac{1±5}{-2} tenglamasini yeching, bunda ± manfiy. 1 dan 5 ni ayirish.

$$x=-\frac{4}{-2}$$

-4 ni -2 ga bo'lish.

$$x=2$$

Tenglama yechildi.

$$x=-3 x=2$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -1,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 6x\left(x+1\right) ga, x+1,x,6 ning eng kichik karralisiga ko‘paytiring.

$$6xx+\left(6x+6\right)\left(x+1\right)=13x\left(x+1\right)$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$6x^{2}+\left(6x+6\right)\left(x+1\right)=13x\left(x+1\right)$$

6x+6 ga x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$6x^{2}+6x^{2}+12x+6=13x\left(x+1\right)$$

12x^{2} ni olish uchun 6x^{2} va 6x^{2} ni birlashtirish.

$$12x^{2}+12x+6=13x\left(x+1\right)$$

13x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$12x^{2}+12x+6=13x^{2}+13x$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$12x^{2}+12x+6-13x^{2}=13x$$

-x^{2} ni olish uchun 12x^{2} va -13x^{2} ni birlashtirish.

$$-x^{2}+12x+6=13x$$

Ikkala tarafdan 13x ni ayirish.

$$-x^{2}+12x+6-13x=0$$

-x ni olish uchun 12x va -13x ni birlashtirish.

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

Ikkala tarafdan 6 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}-x}{-1}=-\frac{6}{-1}$$

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

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

-1 ni -1 ga bo'lish.

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

-6 ni -1 ga bo'lish.

$$x^{2}+x=6$$

1 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{2} olish uchun. Keyin, \frac{1}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$x^{2}+x+\frac{1}{4}=6+\frac{1}{4}$$

6 ni \frac{1}{4} ga qo'shish.

$$x^{2}+x+\frac{1}{4}=\frac{25}{4}$$

x^{2}+x+\frac{1}{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.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{5}{2} x+\frac{1}{2}=-\frac{5}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=2 x=-3$$