3x^2-8x+4/x-2=5x+8 eching

$$\frac { 3 x ^ { 2 } - 8 x + 4 } { x - 2 } = 5 x + 8$$

x uchun yechish

$$x=-5$$

Guruhlash orqali faktorlash qadamlari

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-2 ga ko'paytirish.

$$3x^{2}-8x+4=5x\left(x-2\right)+\left(x-2\right)\times 8$$

5x ga x-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}-8x+4=5x^{2}-10x+\left(x-2\right)\times 8$$

x-2 ga 8 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}-8x+4=5x^{2}-10x+8x-16$$

-2x ni olish uchun -10x va 8x ni birlashtirish.

$$3x^{2}-8x+4=5x^{2}-2x-16$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$3x^{2}-8x+4-5x^{2}=-2x-16$$

-2x^{2} ni olish uchun 3x^{2} va -5x^{2} ni birlashtirish.

$$-2x^{2}-8x+4=-2x-16$$

2x ni ikki tarafga qo’shing.

$$-2x^{2}-8x+4+2x=-16$$

-6x ni olish uchun -8x va 2x ni birlashtirish.

$$-2x^{2}-6x+4=-16$$

16 ni ikki tarafga qo’shing.

$$-2x^{2}-6x+4+16=0$$

20 olish uchun 4 va 16'ni qo'shing.

$$-2x^{2}-6x+20=0$$

Ikki tarafini 2 ga bo‘ling.

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

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -x^{2}+ax+bx+10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=-10=-10$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -10-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-10 2,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-10=-9 2-5=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=2 b=-5$$

-x^{2}-3x+10 ni \left(-x^{2}+2x\right)+\left(-5x+10\right) sifatida qaytadan yozish.

$$\left(-x^{2}+2x\right)+\left(-5x+10\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$x\left(-x+2\right)+5\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(x+5\right)$$

Tenglamani yechish uchun -x+2=0 va x+5=0 ni yeching.

$$x=2 x=-5$$

x qiymati 2 teng bo‘lmaydi.

$$x=-5$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-2 ga ko'paytirish.

$$3x^{2}-8x+4=5x\left(x-2\right)+\left(x-2\right)\times 8$$

5x ga x-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}-8x+4=5x^{2}-10x+\left(x-2\right)\times 8$$

x-2 ga 8 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}-8x+4=5x^{2}-10x+8x-16$$

-2x ni olish uchun -10x va 8x ni birlashtirish.

$$3x^{2}-8x+4=5x^{2}-2x-16$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$3x^{2}-8x+4-5x^{2}=-2x-16$$

-2x^{2} ni olish uchun 3x^{2} va -5x^{2} ni birlashtirish.

$$-2x^{2}-8x+4=-2x-16$$

2x ni ikki tarafga qo’shing.

$$-2x^{2}-8x+4+2x=-16$$

-6x ni olish uchun -8x va 2x ni birlashtirish.

$$-2x^{2}-6x+4=-16$$

16 ni ikki tarafga qo’shing.

$$-2x^{2}-6x+4+16=0$$

20 olish uchun 4 va 16'ni qo'shing.

$$-2x^{2}-6x+20=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -2 ni a, -6 ni b va 20 ni c bilan almashtiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+8\times 20}}{2\left(-2\right)}$$

8 ni 20 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+160}}{2\left(-2\right)}$$

36 ni 160 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{196}}{2\left(-2\right)}$$

196 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±14}{2\left(-2\right)}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±14}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{6±14}{-4}$$

x=\frac{6±14}{-4} tenglamasini yeching, bunda ± musbat. 6 ni 14 ga qo'shish.

$$x=\frac{20}{-4}$$

20 ni -4 ga bo'lish.

$$x=-5$$

x=\frac{6±14}{-4} tenglamasini yeching, bunda ± manfiy. 6 dan 14 ni ayirish.

$$x=-\frac{8}{-4}$$

-8 ni -4 ga bo'lish.

$$x=2$$

Tenglama yechildi.

$$x=-5 x=2$$

x qiymati 2 teng bo‘lmaydi.

$$x=-5$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-2 ga ko'paytirish.

$$3x^{2}-8x+4=5x\left(x-2\right)+\left(x-2\right)\times 8$$

5x ga x-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}-8x+4=5x^{2}-10x+\left(x-2\right)\times 8$$

x-2 ga 8 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$3x^{2}-8x+4=5x^{2}-10x+8x-16$$

-2x ni olish uchun -10x va 8x ni birlashtirish.

$$3x^{2}-8x+4=5x^{2}-2x-16$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$3x^{2}-8x+4-5x^{2}=-2x-16$$

-2x^{2} ni olish uchun 3x^{2} va -5x^{2} ni birlashtirish.

$$-2x^{2}-8x+4=-2x-16$$

2x ni ikki tarafga qo’shing.

$$-2x^{2}-8x+4+2x=-16$$

-6x ni olish uchun -8x va 2x ni birlashtirish.

$$-2x^{2}-6x+4=-16$$

Ikkala tarafdan 4 ni ayirish.

$$-2x^{2}-6x=-16-4$$

-20 olish uchun -16 dan 4 ni ayirish.

$$-2x^{2}-6x=-20$$

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}-6x}{-2}=-\frac{20}{-2}$$

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

$$x^{2}+\left(-\frac{6}{-2}\right)x=-\frac{20}{-2}$$

-6 ni -2 ga bo'lish.

$$x^{2}+3x=-\frac{20}{-2}$$

-20 ni -2 ga bo'lish.

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

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=10+\frac{9}{4}$$

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

$$x^{2}+3x+\frac{9}{4}=\frac{49}{4}$$

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.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Qisqartirish.

$$x+\frac{3}{2}=\frac{7}{2} x+\frac{3}{2}=-\frac{7}{2}$$

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

$$x=2 x=-5$$

x qiymati 2 teng bo‘lmaydi.

$$x=-5$$