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

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

x uchun yechish

$$x=-4$$

$$x=1$$

Guruhlash orqali faktorlash qadamlari

x qiymati -2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 3\left(x+2\right) ga, x+2,3 ning eng kichik karralisiga ko‘paytiring.

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

6 hosil qilish uchun 3 va 2 ni ko'paytirish.

$$6+3\left(x+2\right)\left(-\frac{1}{3}\right)=\left(x+2\right)x$$

-1 hosil qilish uchun 3 va -\frac{1}{3} ni ko'paytirish.

$$6-\left(x+2\right)=\left(x+2\right)x$$

x+2 teskarisini topish uchun har birining teskarisini toping.

$$6-x-2=\left(x+2\right)x$$

4 olish uchun 6 dan 2 ni ayirish.

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

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

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

Ikkala tarafdan x^{2} ni ayirish.

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

Ikkala tarafdan 2x ni ayirish.

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

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

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

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

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

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

$$a+b=-3 ab=-4=-4$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -4-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-4 2,-2$$

Har bir juftlik yigʻindisini hisoblang.

$$1-4=-3 2-2=0$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=1 b=-4$$

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

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

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

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

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

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

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

$$x=1 x=-4$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 3\left(x+2\right) ga, x+2,3 ning eng kichik karralisiga ko‘paytiring.

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

6 hosil qilish uchun 3 va 2 ni ko'paytirish.

$$6+3\left(x+2\right)\left(-\frac{1}{3}\right)=\left(x+2\right)x$$

-1 hosil qilish uchun 3 va -\frac{1}{3} ni ko'paytirish.

$$6-\left(x+2\right)=\left(x+2\right)x$$

x+2 teskarisini topish uchun har birining teskarisini toping.

$$6-x-2=\left(x+2\right)x$$

4 olish uchun 6 dan 2 ni ayirish.

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

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

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

Ikkala tarafdan x^{2} ni ayirish.

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

Ikkala tarafdan 2x ni ayirish.

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

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

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

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.

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

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

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

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-1\right)}$$

9 ni 16 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-1\right)}$$

25 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±5}{2\left(-1\right)}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±5}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{3±5}{-2}$$

x=\frac{3±5}{-2} tenglamasini yeching, bunda ± musbat. 3 ni 5 ga qo'shish.

$$x=\frac{8}{-2}$$

8 ni -2 ga bo'lish.

$$x=-4$$

x=\frac{3±5}{-2} tenglamasini yeching, bunda ± manfiy. 3 dan 5 ni ayirish.

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

-2 ni -2 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=-4 x=1$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 3\left(x+2\right) ga, x+2,3 ning eng kichik karralisiga ko‘paytiring.

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

6 hosil qilish uchun 3 va 2 ni ko'paytirish.

$$6+3\left(x+2\right)\left(-\frac{1}{3}\right)=\left(x+2\right)x$$

-1 hosil qilish uchun 3 va -\frac{1}{3} ni ko'paytirish.

$$6-\left(x+2\right)=\left(x+2\right)x$$

x+2 teskarisini topish uchun har birining teskarisini toping.

$$6-x-2=\left(x+2\right)x$$

4 olish uchun 6 dan 2 ni ayirish.

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

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

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

Ikkala tarafdan x^{2} ni ayirish.

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

Ikkala tarafdan 2x ni ayirish.

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

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

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

Ikkala tarafdan 4 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

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

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

Ikki tarafini -1 ga bo‘ling.

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

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

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

-3 ni -1 ga bo'lish.

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

-4 ni -1 ga bo'lish.

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

3 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{2} olish uchun. Keyin, \frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{2} kvadratini chiqarish.

$$x^{2}+3x+\frac{9}{4}=4+\frac{9}{4}$$

4 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=\frac{25}{4}$$

x^{2}+3x+\frac{9}{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.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{3}{2} ni ayirish.

$$x=1 x=-4$$