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

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

x uchun yechish

$$x = \frac{\sqrt{41} + 3}{2} \approx 4,701562119$$

$$x=\frac{3-\sqrt{41}}{2}\approx -1,701562119$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -3,-2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x+2\right)\left(x+3\right) ga, x+3,x^{2}+5x+6 ning eng kichik karralisiga ko‘paytiring.

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

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

$$x^{2}-2x-8=1x$$

Ikkala tarafdan 1x ni ayirish.

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

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

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

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 -8 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-8\right)}}{2}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-8\right)}}{2}$$

-4 ni -8 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+32}}{2}$$

9 ni 32 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{41}}{2}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{41}}{2}$$

x=\frac{3±\sqrt{41}}{2} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}+3}{2}$$

x=\frac{3±\sqrt{41}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{41} ni ayirish.

$$x=\frac{3-\sqrt{41}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}+3}{2} x=\frac{3-\sqrt{41}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -3,-2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x+2\right)\left(x+3\right) ga, x+3,x^{2}+5x+6 ning eng kichik karralisiga ko‘paytiring.

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

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

$$x^{2}-2x-8=1x$$

Ikkala tarafdan 1x ni ayirish.

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

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

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

8 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

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

-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.

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

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

$$x^{2}-3x+\frac{9}{4}=8+\frac{9}{4}$$

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

$$x^{2}-3x+\frac{9}{4}=\frac{41}{4}$$

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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{41}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{41}}{2} x-\frac{3}{2}=-\frac{\sqrt{41}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{41}+3}{2} x=\frac{3-\sqrt{41}}{2}$$