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

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

x uchun yechish

$$x=\sqrt{10}+1\approx 4,16227766$$

$$x=1-\sqrt{10}\approx -2,16227766$$

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.

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

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

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

Ikkala tarafdan 1 ni ayirish.

$$x^{2}-2x-8-1=0$$

-9 olish uchun -8 dan 1 ni ayirish.

$$x^{2}-2x-9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -2 ni b va -9 ni c bilan almashtiring.

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

-2 kvadratini chiqarish.

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

-4 ni -9 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+36}}{2}$$

4 ni 36 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{40}}{2}$$

40 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±2\sqrt{10}}{2}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{10}}{2}$$

x=\frac{2±2\sqrt{10}}{2} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{10} ga qo'shish.

$$x=\frac{2\sqrt{10}+2}{2}$$

2+2\sqrt{10} ni 2 ga bo'lish.

$$x=\sqrt{10}+1$$

x=\frac{2±2\sqrt{10}}{2} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{10} ni ayirish.

$$x=\frac{2-2\sqrt{10}}{2}$$

2-2\sqrt{10} ni 2 ga bo'lish.

$$x=1-\sqrt{10}$$

Tenglama yechildi.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$

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.

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

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

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

8 ni ikki tarafga qo’shing.

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

9 olish uchun 1 va 8'ni qo'shing.

$$x^{2}-2x=9$$

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

$$x^{2}-2x+1=9+1$$

9 ni 1 ga qo'shish.

$$x^{2}-2x+1=10$$

x^{2}-2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-1\right)^{2}=10$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{10}$$

Qisqartirish.

$$x-1=\sqrt{10} x-1=-\sqrt{10}$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$