x-1=sqrt{2x} eching

$$x-1= \sqrt{ 2x }$$

x uchun yechish

$$x=\sqrt{3}+2\approx 3,732050808$$

Yechim qadamlari

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-1\right)^{2} kengaytirilishi uchun ishlating.

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

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

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

Ikkala tarafdan 2x ni ayirish.

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

-4x ni olish uchun -2x va -2x ni birlashtirish.

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

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.

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

-4 kvadratini chiqarish.

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

16 ni -4 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{12}}{2}$$

12 ning kvadrat ildizini chiqarish.

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

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±2\sqrt{3}}{2}$$

x=\frac{4±2\sqrt{3}}{2} tenglamasini yeching, bunda ± musbat. 4 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}+4}{2}$$

4+2\sqrt{3} ni 2 ga bo'lish.

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

x=\frac{4±2\sqrt{3}}{2} tenglamasini yeching, bunda ± manfiy. 4 dan 2\sqrt{3} ni ayirish.

$$x=\frac{4-2\sqrt{3}}{2}$$

4-2\sqrt{3} ni 2 ga bo'lish.

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

Tenglama yechildi.

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

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

$$\sqrt{3}+2-1=\sqrt{2\left(\sqrt{3}+2\right)}$$

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

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

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

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

Qisqartirish. x=2-\sqrt{3} qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

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

x-1=\sqrt{2x} tenglamasi noyob yechimga ega.

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