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

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

x uchun yechish

$$x=-2$$

Faktorlash qadamlari

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

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

x+1 ga 2x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

-3 olish uchun -1 dan 2 ni ayirish.

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

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

Ikkala tarafdan x^{2} ni ayirish.

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

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

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

1 ni ikki tarafga qo’shing.

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

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

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

Bu tenglamani yechish uchun x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right) formulasi yordamida x^{2}+x-2 ni faktorlang. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-1 b=2$$

Faktorlangan \left(x+a\right)\left(x+b\right) ifodani olingan qiymatlar bilan qaytadan yozing.

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

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

$$x=1 x=-2$$

x qiymati 1 teng bo‘lmaydi.

$$x=-2$$

Guruhlash orqali faktorlash qadamlari

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

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

x+1 ga 2x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

-3 olish uchun -1 dan 2 ni ayirish.

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

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

Ikkala tarafdan x^{2} ni ayirish.

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

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

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

1 ni ikki tarafga qo’shing.

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

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

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

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

$$a+b=1 ab=1\left(-2\right)=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-1 b=2$$

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

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

Birinchi guruhda x ni va ikkinchi guruhda 2 ni faktordan chiqaring.

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

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

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

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

$$x=1 x=-2$$

x qiymati 1 teng bo‘lmaydi.

$$x=-2$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

x+1 ga 2x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

-3 olish uchun -1 dan 2 ni ayirish.

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

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

Ikkala tarafdan x^{2} ni ayirish.

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

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

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

1 ni ikki tarafga qo’shing.

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

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

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

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

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

1 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+8}}{2}$$

1 ni 8 ga qo'shish.

$$x=\frac{-1±\sqrt{9}}{2}$$

9 ning kvadrat ildizini chiqarish.

$$x=\frac{-1±3}{2}$$

x=\frac{-1±3}{2} tenglamasini yeching, bunda ± musbat. -1 ni 3 ga qo'shish.

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

2 ni 2 ga bo'lish.

$$x=1$$

x=\frac{-1±3}{2} tenglamasini yeching, bunda ± manfiy. -1 dan 3 ni ayirish.

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

-4 ni 2 ga bo'lish.

$$x=-2$$

Tenglama yechildi.

$$x=1 x=-2$$

x qiymati 1 teng bo‘lmaydi.

$$x=-2$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

x+1 ga 2x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

-3 olish uchun -1 dan 2 ni ayirish.

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

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

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

Ikkala tarafdan x^{2} ni ayirish.

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

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

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

3 ni ikki tarafga qo’shing.

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

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

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

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

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

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

2 ni \frac{1}{4} ga qo'shish.

$$x^{2}+x+\frac{1}{4}=\frac{9}{4}$$

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

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=1 x=-2$$

x qiymati 1 teng bo‘lmaydi.

$$x=-2$$