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

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

x uchun yechish

$$x=-2$$

$$x=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

x+4 ga -2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-2x-8+x-2=\left(x-2\right)\left(x+4\right)$$

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

$$-x-8-2=\left(x-2\right)\left(x+4\right)$$

-10 olish uchun -8 dan 2 ni ayirish.

$$-x-10=\left(x-2\right)\left(x+4\right)$$

x-2 ga x+4 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$-x-10=x^{2}+2x-8$$

Ikkala tarafdan x^{2} ni ayirish.

$$-x-10-x^{2}=2x-8$$

Ikkala tarafdan 2x ni ayirish.

$$-x-10-x^{2}-2x=-8$$

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

$$-3x-10-x^{2}=-8$$

8 ni ikki tarafga qo’shing.

$$-3x-10-x^{2}+8=0$$

-2 olish uchun -10 va 8'ni qo'shing.

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

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.

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

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 -2 ni c bilan almashtiring.

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

-3 kvadratini chiqarish.

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

-4 ni -1 marotabaga ko'paytirish.

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

4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2\left(-1\right)}$$

9 ni -8 ga qo'shish.

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

1 ning kvadrat ildizini chiqarish.

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

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±1}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{3±1}{-2}$$

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

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

4 ni -2 ga bo'lish.

$$x=-2$$

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

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

2 ni -2 ga bo'lish.

$$x=-1$$

Tenglama yechildi.

$$x=-2 x=-1$$

Kvadratni yakunlash bo'yicha qadamlar

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

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

x+4 ga -2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-2x-8+x-2=\left(x-2\right)\left(x+4\right)$$

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

$$-x-8-2=\left(x-2\right)\left(x+4\right)$$

-10 olish uchun -8 dan 2 ni ayirish.

$$-x-10=\left(x-2\right)\left(x+4\right)$$

x-2 ga x+4 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$-x-10=x^{2}+2x-8$$

Ikkala tarafdan x^{2} ni ayirish.

$$-x-10-x^{2}=2x-8$$

Ikkala tarafdan 2x ni ayirish.

$$-x-10-x^{2}-2x=-8$$

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

$$-3x-10-x^{2}=-8$$

10 ni ikki tarafga qo’shing.

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

2 olish uchun -8 va 10'ni qo'shing.

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

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

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

Ikki tarafini -1 ga bo‘ling.

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

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

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

-3 ni -1 ga bo'lish.

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

2 ni -1 ga bo'lish.

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

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.

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

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

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

-2 ni \frac{9}{4} ga qo'shish.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{4}$$

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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

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

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

$$x=-1 x=-2$$