6x-4/x-1+5x-2=3x+1 eching

$$\frac { 6 x - 4 } { x - 1 } + 5 x - 2 = 3 x + 1$$

x uchun yechish

$$x=-1$$

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

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 1 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-1 ga ko'paytirish.

$$6x-4+5x\left(x-1\right)+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

5x ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$6x-4+5x^{2}-5x+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

x ni olish uchun 6x va -5x ni birlashtirish.

$$x-4+5x^{2}+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

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

$$x-4+5x^{2}-2x+2=3x\left(x-1\right)+x-1$$

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

$$-x-4+5x^{2}+2=3x\left(x-1\right)+x-1$$

-2 olish uchun -4 va 2'ni qo'shing.

$$-x-2+5x^{2}=3x\left(x-1\right)+x-1$$

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

$$-x-2+5x^{2}=3x^{2}-3x+x-1$$

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

$$-x-2+5x^{2}=3x^{2}-2x-1$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$-x-2+5x^{2}-3x^{2}=-2x-1$$

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

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

2x ni ikki tarafga qo’shing.

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

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

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

1 ni ikki tarafga qo’shing.

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

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

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

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.

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

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

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

1 kvadratini chiqarish.

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

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+8}}{2\times 2}$$

1 ni 8 ga qo'shish.

$$x=\frac{-1±\sqrt{9}}{2\times 2}$$

9 ning kvadrat ildizini chiqarish.

$$x=\frac{-1±3}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±3}{4}$$

x=\frac{-1±3}{4} tenglamasini yeching, bunda ± musbat. -1 ni 3 ga qo'shish.

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

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

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

x=\frac{-1±3}{4} tenglamasini yeching, bunda ± manfiy. -1 dan 3 ni ayirish.

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

-4 ni 4 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 1 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-1 ga ko'paytirish.

$$6x-4+5x\left(x-1\right)+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

5x ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$6x-4+5x^{2}-5x+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

x ni olish uchun 6x va -5x ni birlashtirish.

$$x-4+5x^{2}+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

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

$$x-4+5x^{2}-2x+2=3x\left(x-1\right)+x-1$$

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

$$-x-4+5x^{2}+2=3x\left(x-1\right)+x-1$$

-2 olish uchun -4 va 2'ni qo'shing.

$$-x-2+5x^{2}=3x\left(x-1\right)+x-1$$

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

$$-x-2+5x^{2}=3x^{2}-3x+x-1$$

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

$$-x-2+5x^{2}=3x^{2}-2x-1$$

Ikkala tarafdan 3x^{2} ni ayirish.

$$-x-2+5x^{2}-3x^{2}=-2x-1$$

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

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

2x ni ikki tarafga qo’shing.

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

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

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

2 ni ikki tarafga qo’shing.

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

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

$$x+2x^{2}=1$$

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

$$2x^{2}+x=1$$

Ikki tarafini 2 ga bo‘ling.

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

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

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

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

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

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