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

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

x uchun yechish

$$x=-\frac{2}{3}\approx -0,666666667$$

$$x=-1$$

Guruhlash orqali faktorlash qadamlari

x qiymati -2,1,2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-2\right)\left(x-1\right)\left(x+2\right) ga, x^{2}+x-2,x^{2}-4,x^{2}-3x+2 ning eng kichik karralisiga ko‘paytiring.

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

\left(x+2\right)^{2} hosil qilish uchun x+2 va x+2 ni ko'paytirish.

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

x-2 ga x^{2}-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

x-1 ga 3x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

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

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

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

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

2 olish uchun 4 dan 2 ni ayirish.

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

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

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

x^{2}-3x+2 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

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

x^{2}+4x+4 teskarisini topish uchun har birining teskarisini toping.

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

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

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

-8x ni olish uchun -4x va -4x ni birlashtirish.

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

0 olish uchun 4 dan 4 ni ayirish.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x$$

Ikkala tarafdan x^{3} ni ayirish.

$$x^{3}-3x+x^{2}+2-x^{3}=-2x^{2}-8x$$

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

$$-3x+x^{2}+2=-2x^{2}-8x$$

2x^{2} ni ikki tarafga qo’shing.

$$-3x+x^{2}+2+2x^{2}=-8x$$

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

$$-3x+3x^{2}+2=-8x$$

8x ni ikki tarafga qo’shing.

$$-3x+3x^{2}+2+8x=0$$

5x ni olish uchun -3x va 8x ni birlashtirish.

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

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

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

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

$$a+b=5 ab=3\times 2=6$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 6-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,6 2,3$$

Har bir juftlik yigʻindisini hisoblang.

$$1+6=7 2+3=5$$

Yechim – 5 yigʻindisini beruvchi juftlik.

$$a=2 b=3$$

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

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

3x^{2}+2x ichida x ni ajrating.

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

Distributiv funktsiyasidan foydalangan holda 3x+2 umumiy terminini chiqaring.

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

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

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

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -2,1,2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-2\right)\left(x-1\right)\left(x+2\right) ga, x^{2}+x-2,x^{2}-4,x^{2}-3x+2 ning eng kichik karralisiga ko‘paytiring.

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

\left(x+2\right)^{2} hosil qilish uchun x+2 va x+2 ni ko'paytirish.

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

x-2 ga x^{2}-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

x-1 ga 3x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

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

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

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

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

2 olish uchun 4 dan 2 ni ayirish.

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

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

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

x^{2}-3x+2 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

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

x^{2}+4x+4 teskarisini topish uchun har birining teskarisini toping.

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

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

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

-8x ni olish uchun -4x va -4x ni birlashtirish.

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

0 olish uchun 4 dan 4 ni ayirish.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x$$

Ikkala tarafdan x^{3} ni ayirish.

$$x^{3}-3x+x^{2}+2-x^{3}=-2x^{2}-8x$$

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

$$-3x+x^{2}+2=-2x^{2}-8x$$

2x^{2} ni ikki tarafga qo’shing.

$$-3x+x^{2}+2+2x^{2}=-8x$$

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

$$-3x+3x^{2}+2=-8x$$

8x ni ikki tarafga qo’shing.

$$-3x+3x^{2}+2+8x=0$$

5x ni olish uchun -3x va 8x ni birlashtirish.

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

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.

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

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

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

5 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-12\times 2}}{2\times 3}$$

-12 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-24}}{2\times 3}$$

25 ni -24 ga qo'shish.

$$x=\frac{-5±\sqrt{1}}{2\times 3}$$

1 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±1}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-5±1}{6}$$

x=\frac{-5±1}{6} tenglamasini yeching, bunda ± musbat. -5 ni 1 ga qo'shish.

$$x=-\frac{4}{6}$$

\frac{-4}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{-5±1}{6} tenglamasini yeching, bunda ± manfiy. -5 dan 1 ni ayirish.

$$x=-\frac{6}{6}$$

-6 ni 6 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -2,1,2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-2\right)\left(x-1\right)\left(x+2\right) ga, x^{2}+x-2,x^{2}-4,x^{2}-3x+2 ning eng kichik karralisiga ko‘paytiring.

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

\left(x+2\right)^{2} hosil qilish uchun x+2 va x+2 ni ko'paytirish.

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

x-2 ga x^{2}-2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

x-1 ga 3x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

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

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

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

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

2 olish uchun 4 dan 2 ni ayirish.

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

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

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

x^{2}-3x+2 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x+2\right)^{2} kengaytirilishi uchun ishlating.

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

x^{2}+4x+4 teskarisini topish uchun har birining teskarisini toping.

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

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

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

-8x ni olish uchun -4x va -4x ni birlashtirish.

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

0 olish uchun 4 dan 4 ni ayirish.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x$$

Ikkala tarafdan x^{3} ni ayirish.

$$x^{3}-3x+x^{2}+2-x^{3}=-2x^{2}-8x$$

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

$$-3x+x^{2}+2=-2x^{2}-8x$$

2x^{2} ni ikki tarafga qo’shing.

$$-3x+x^{2}+2+2x^{2}=-8x$$

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

$$-3x+3x^{2}+2=-8x$$

8x ni ikki tarafga qo’shing.

$$-3x+3x^{2}+2+8x=0$$

5x ni olish uchun -3x va 8x ni birlashtirish.

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

Ikkala tarafdan 2 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

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

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

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

Ikki tarafini 3 ga bo‘ling.

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

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

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

\frac{5}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{6} olish uchun. Keyin, \frac{5}{6} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{6} kvadratini chiqarish.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{2}{3}+\frac{25}{36}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{2}{3} ni \frac{25}{36} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{1}{36}$$

x^{2}+\frac{5}{3}x+\frac{25}{36} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{1}{36}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{5}{6} ni ayirish.

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