(1+sqrt{2})^4=a+bsqrt{2} eching

$$( 1 + \sqrt { 2 } ) ^ { 4 } = a + b \sqrt { 2 }$$

a uchun yechish

$$a=\sqrt{2}\left(12-b\right)+17$$

Yechim qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$a+b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}$$

Ikkala tarafdan b\sqrt{2} ni ayirish.

$$a=\left(1+\sqrt{2}\right)^{4}-b\sqrt{2}$$

Shartlarni qayta saralash.

$$a=-\sqrt{2}b+\left(\sqrt{2}+1\right)^{4}$$

b uchun yechish

$$b=-\frac{\sqrt{2}\left(a-12\sqrt{2}-17\right)}{2}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$a+b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}$$

Ikkala tarafdan a ni ayirish.

$$b\sqrt{2}=\left(1+\sqrt{2}\right)^{4}-a$$

Tenglama standart shaklda.

$$\sqrt{2}b=-a+\left(\sqrt{2}+1\right)^{4}$$

Ikki tarafini \sqrt{2} ga bo‘ling.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+12\sqrt{2}+17}{\sqrt{2}}$$

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

$$b=\frac{-a+12\sqrt{2}+17}{\sqrt{2}}$$

17+12\sqrt{2}-a ni \sqrt{2} ga bo'lish.

$$b=\frac{\sqrt{2}\left(-a+12\sqrt{2}+17\right)}{2}$$