-4/x+1-3/x=-2 eching

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

x uchun yechish

$$x=-\frac{1}{2}=-0,5$$

$$x=3$$

Guruhlash orqali faktorlash qadamlari

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

$$-x\times 4-\left(x+1\right)\times 3=-2x\left(x+1\right)$$

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

$$-x\times 4-\left(3x+3\right)=-2x\left(x+1\right)$$

3x+3 teskarisini topish uchun har birining teskarisini toping.

$$-x\times 4-3x-3=-2x\left(x+1\right)$$

-2x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

2x^{2} ni ikki tarafga qo’shing.

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

2x ni ikki tarafga qo’shing.

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

-x ni olish uchun -3x va 2x ni birlashtirish.

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

-4 hosil qilish uchun -1 va 4 ni ko'paytirish.

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

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

$$-5x-3+2x^{2}=0$$

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

$$2x^{2}-5x-3=0$$

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

$$a+b=-5 ab=2\left(-3\right)=-6$$

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 – -5 yigʻindisini beruvchi juftlik.

$$a=-6 b=1$$

2x^{2}-5x-3 ni \left(2x^{2}-6x\right)+\left(x-3\right) sifatida qaytadan yozish.

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

2x^{2}-6x ichida 2x ni ajrating.

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

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(2x+1\right)$$

Tenglamani yechish uchun x-3=0 va 2x+1=0 ni yeching.

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

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$-x\times 4-\left(x+1\right)\times 3=-2x\left(x+1\right)$$

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

$$-x\times 4-\left(3x+3\right)=-2x\left(x+1\right)$$

3x+3 teskarisini topish uchun har birining teskarisini toping.

$$-x\times 4-3x-3=-2x\left(x+1\right)$$

-2x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

2x^{2} ni ikki tarafga qo’shing.

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

2x ni ikki tarafga qo’shing.

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

-x ni olish uchun -3x va 2x ni birlashtirish.

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

-4 hosil qilish uchun -1 va 4 ni ko'paytirish.

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

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

$$-5x-3+2x^{2}=0$$

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.

$$2x^{2}-5x-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -5 ni b va -3 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-3\right)}}{2\times 2}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-3\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-3\right)}}{2\times 2}$$

-8 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 2}$$

25 ni 24 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\times 2}$$

49 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±7}{2\times 2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±7}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{5±7}{4}$$

x=\frac{5±7}{4} tenglamasini yeching, bunda ± musbat. 5 ni 7 ga qo'shish.

$$x=\frac{12}{4}$$

12 ni 4 ga bo'lish.

$$x=3$$

x=\frac{5±7}{4} tenglamasini yeching, bunda ± manfiy. 5 dan 7 ni ayirish.

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

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

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

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

$$-x\times 4-\left(x+1\right)\times 3=-2x\left(x+1\right)$$

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

$$-x\times 4-\left(3x+3\right)=-2x\left(x+1\right)$$

3x+3 teskarisini topish uchun har birining teskarisini toping.

$$-x\times 4-3x-3=-2x\left(x+1\right)$$

-2x ga x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

2x^{2} ni ikki tarafga qo’shing.

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

2x ni ikki tarafga qo’shing.

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

-x ni olish uchun -3x va 2x ni birlashtirish.

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

3 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

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

-4 hosil qilish uchun -1 va 4 ni ko'paytirish.

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

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

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

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

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

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-5x}{2}=\frac{3}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}-\frac{5}{2}x=\frac{3}{2}$$

-\frac{5}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{4} olish uchun. Keyin, -\frac{5}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\frac{3}{2}+\left(-\frac{5}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{4} kvadratini chiqarish.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{2} ni \frac{25}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{49}{16}$$

x^{2}-\frac{5}{2}x+\frac{25}{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.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Qisqartirish.

$$x-\frac{5}{4}=\frac{7}{4} x-\frac{5}{4}=-\frac{7}{4}$$

\frac{5}{4} ni tenglamaning ikkala tarafiga qo'shish.

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