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

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

x uchun yechish

$$x=-\frac{3}{4}=-0,75$$

$$x=1$$

Guruhlash orqali faktorlash qadamlari

x qiymati -1,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2x\left(x+1\right) ga, 2x,x+1,x\left(x+1\right) ning eng kichik karralisiga ko‘paytiring.

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

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

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

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 1 va 2 ni qo‘shib, 3 ni oling.

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

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

$$2x^{3}+x+2x^{2}+1-2x^{3}=6x^{2}-2$$

Ikkala tarafdan 6x^{2} ni ayirish.

$$2x^{3}+x+2x^{2}+1-2x^{3}-6x^{2}=-2$$

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

$$2x^{3}+x-4x^{2}+1-2x^{3}=-2$$

2 ni ikki tarafga qo’shing.

$$2x^{3}+x-4x^{2}+1-2x^{3}+2=0$$

3 olish uchun 1 va 2'ni qo'shing.

$$2x^{3}+x-4x^{2}+3-2x^{3}=0$$

0 ni olish uchun 2x^{3} va -2x^{3} ni birlashtirish.

$$x-4x^{2}+3=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$-4x^{2}+x+3=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -4x^{2}+ax+bx+3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=-4\times 3=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,12 -2,6 -3,4$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+12=11 -2+6=4 -3+4=1$$

Yechim – 1 yigʻindisini beruvchi juftlik.

$$a=4 b=-3$$

-4x^{2}+x+3 ni \left(-4x^{2}+4x\right)+\left(-3x+3\right) sifatida qaytadan yozish.

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

Birinchi guruhda 4x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$4x\left(-x+1\right)+3\left(-x+1\right)$$

Distributiv funktsiyasidan foydalangan holda -x+1 umumiy terminini chiqaring.

$$\left(-x+1\right)\left(4x+3\right)$$

Tenglamani yechish uchun -x+1=0 va 4x+3=0 ni yeching.

$$x=1 x=-\frac{3}{4}$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -1,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2x\left(x+1\right) ga, 2x,x+1,x\left(x+1\right) ning eng kichik karralisiga ko‘paytiring.

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

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

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

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 1 va 2 ni qo‘shib, 3 ni oling.

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

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

$$2x^{3}+x+2x^{2}+1-2x^{3}=6x^{2}-2$$

Ikkala tarafdan 6x^{2} ni ayirish.

$$2x^{3}+x+2x^{2}+1-2x^{3}-6x^{2}=-2$$

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

$$2x^{3}+x-4x^{2}+1-2x^{3}=-2$$

2 ni ikki tarafga qo’shing.

$$2x^{3}+x-4x^{2}+1-2x^{3}+2=0$$

3 olish uchun 1 va 2'ni qo'shing.

$$2x^{3}+x-4x^{2}+3-2x^{3}=0$$

0 ni olish uchun 2x^{3} va -2x^{3} ni birlashtirish.

$$x-4x^{2}+3=0$$

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.

$$-4x^{2}+x+3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -4 ni a, 1 ni b va 3 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+16\times 3}}{2\left(-4\right)}$$

16 ni 3 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+48}}{2\left(-4\right)}$$

1 ni 48 ga qo'shish.

$$x=\frac{-1±\sqrt{49}}{2\left(-4\right)}$$

49 ning kvadrat ildizini chiqarish.

$$x=\frac{-1±7}{2\left(-4\right)}$$

2 ni -4 marotabaga ko'paytirish.

$$x=\frac{-1±7}{-8}$$

x=\frac{-1±7}{-8} tenglamasini yeching, bunda ± musbat. -1 ni 7 ga qo'shish.

$$x=\frac{6}{-8}$$

\frac{6}{-8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{3}{4}$$

x=\frac{-1±7}{-8} tenglamasini yeching, bunda ± manfiy. -1 dan 7 ni ayirish.

$$x=-\frac{8}{-8}$$

-8 ni -8 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=-\frac{3}{4} x=1$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -1,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2x\left(x+1\right) ga, 2x,x+1,x\left(x+1\right) ning eng kichik karralisiga ko‘paytiring.

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

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

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

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 1 va 2 ni qo‘shib, 3 ni oling.

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

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

$$2x^{3}+x+2x^{2}+1-2x^{3}=6x^{2}-2$$

Ikkala tarafdan 6x^{2} ni ayirish.

$$2x^{3}+x+2x^{2}+1-2x^{3}-6x^{2}=-2$$

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

$$2x^{3}+x-4x^{2}+1-2x^{3}=-2$$

Ikkala tarafdan 1 ni ayirish.

$$2x^{3}+x-4x^{2}-2x^{3}=-2-1$$

-3 olish uchun -2 dan 1 ni ayirish.

$$2x^{3}+x-4x^{2}-2x^{3}=-3$$

0 ni olish uchun 2x^{3} va -2x^{3} ni birlashtirish.

$$x-4x^{2}=-3$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-4x^{2}+x=-3$$

Ikki tarafini -4 ga bo‘ling.

$$\frac{-4x^{2}+x}{-4}=-\frac{3}{-4}$$

-4 ga bo'lish -4 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{1}{-4}x=-\frac{3}{-4}$$

1 ni -4 ga bo'lish.

$$x^{2}-\frac{1}{4}x=-\frac{3}{-4}$$

-3 ni -4 ga bo'lish.

$$x^{2}-\frac{1}{4}x=\frac{3}{4}$$

-\frac{1}{4} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{8} olish uchun. Keyin, -\frac{1}{8} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{8}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{8} kvadratini chiqarish.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{3}{4}+\frac{1}{64}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{4} ni \frac{1}{64} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{49}{64}$$

x^{2}-\frac{1}{4}x+\frac{1}{64} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{49}{64}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Qisqartirish.

$$x-\frac{1}{8}=\frac{7}{8} x-\frac{1}{8}=-\frac{7}{8}$$

\frac{1}{8} ni tenglamaning ikkala tarafiga qo'shish.

$$x=1 x=-\frac{3}{4}$$