x^2-4x-1/(x^2+1)^2=0 eching

$$\frac { x ^ { 2 } - 4 x - 1 } { ( x ^ { 2 } + 1 ) ^ { 2 } } = 0$$

x uchun yechish

$$x=\sqrt{5}+2\approx 4,236067977$$

$$x=2-\sqrt{5}\approx -0,236067977$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tenglamaning ikkala tarafini \left(x^{2}+1\right)^{2} ga ko'paytirish.

$$x^{2}-4x-1=0$$

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

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

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)}}{2}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+4}}{2}$$

16 ni 4 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{20}}{2}$$

20 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-4\right)±2\sqrt{5}}{2}$$

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±2\sqrt{5}}{2}$$

x=\frac{4±2\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 4 ni 2\sqrt{5} ga qo'shish.

$$x=\frac{2\sqrt{5}+4}{2}$$

4+2\sqrt{5} ni 2 ga bo'lish.

$$x=\sqrt{5}+2$$

x=\frac{4±2\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 4 dan 2\sqrt{5} ni ayirish.

$$x=\frac{4-2\sqrt{5}}{2}$$

4-2\sqrt{5} ni 2 ga bo'lish.

$$x=2-\sqrt{5}$$

Tenglama yechildi.

$$x=\sqrt{5}+2 x=2-\sqrt{5}$$

Kvadratni yakunlash bo'yicha qadamlar

Tenglamaning ikkala tarafini \left(x^{2}+1\right)^{2} ga ko'paytirish.

$$x^{2}-4x-1=0$$

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

$$x^{2}-4x=1$$

-4 ni bo‘lish, x shartining koeffitsienti, 2 ga -2 olish uchun. Keyin, -2 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

-2 kvadratini chiqarish.

$$x^{2}-4x+4=1+4$$

1 ni 4 ga qo'shish.

$$x^{2}-4x+4=5$$

x^{2}-4x+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.

$$\left(x-2\right)^{2}=5$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{5}$$

Qisqartirish.

$$x-2=\sqrt{5} x-2=-\sqrt{5}$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{5}+2 x=2-\sqrt{5}$$