1+x/sqrt{2}=1 eching

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

x uchun yechish

$$x=\sqrt{2}-1\approx 0,414213562$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\frac{1+x}{\sqrt{2}} maxrajini \sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

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

\sqrt{2} kvadrati – 2.

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

1+x ga \sqrt{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Ikkala tarafini 2 ga ko‘paytiring.

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

Ikkala tarafdan \sqrt{2} ni ayirish.

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

Tenglama standart shaklda.

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

Ikki tarafini \sqrt{2} ga bo‘ling.

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

\sqrt{2} ga bo'lish \sqrt{2} ga ko'paytirishni bekor qiladi.

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

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

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