(x)=sqrt{(x-1)} eching

$$( x ) = \sqrt { ( x - 1 ) }$$

x uchun yechish (complex solution)

$$x=\frac{1+\sqrt{3}i}{2}\approx 0,5+0,866025404i$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{x-1} ga hisoblang va x-1 ni qiymatni oling.

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

Ikkala tarafdan x ni ayirish.

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

1 ni ikki tarafga qo’shing.

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

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

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

1 ni -4 ga qo'shish.

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

-3 ning kvadrat ildizini chiqarish.

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

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{3}i}{2}$$

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

$$x=\frac{1+\sqrt{3}i}{2}$$

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

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

Tenglama yechildi.

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

x=\sqrt{x-1} tenglamasida x uchun \frac{1+\sqrt{3}i}{2} ni almashtiring.

$$\frac{1+\sqrt{3}i}{2}=\sqrt{\frac{1+\sqrt{3}i}{2}-1}$$

Qisqartirish. x=\frac{1+\sqrt{3}i}{2} tenglamani qoniqtiradi.

$$\frac{1}{2}+\frac{1}{2}i\times 3^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}i\times 3^{\frac{1}{2}}$$

x=\sqrt{x-1} tenglamasida x uchun \frac{-\sqrt{3}i+1}{2} ni almashtiring.

$$\frac{-\sqrt{3}i+1}{2}=\sqrt{\frac{-\sqrt{3}i+1}{2}-1}$$

Qisqartirish. x=\frac{-\sqrt{3}i+1}{2} qiymati bu tenglamani qoniqtirmaydi.

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

x=\sqrt{x-1} tenglamasi noyob yechimga ega.

$$x=\frac{1+\sqrt{3}i}{2}$$