frac{10-sqrt{18}}{sqrt{2}}=a+bsqrt{2} eching

$$\frac { 10 - \sqrt { 18 } } { \sqrt { 2 } } = a + b \sqrt { 2 }$$

b uchun yechish

$$b=-\frac{\sqrt{2}\left(a+3-5\sqrt{2}\right)}{2}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Faktor: 18=3^{2}\times 2. \sqrt{3^{2}\times 2} koʻpaytmasining kvadrat ildizini \sqrt{3^{2}}\sqrt{2} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing. 3^{2} ning kvadrat ildizini chiqarish.

$$\frac{10-3\sqrt{2}}{\sqrt{2}}=a+b\sqrt{2}$$

\frac{10-3\sqrt{2}}{\sqrt{2}} maxrajini \sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$\frac{\left(10-3\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}=a+b\sqrt{2}$$

\sqrt{2} kvadrati – 2.

$$\frac{\left(10-3\sqrt{2}\right)\sqrt{2}}{2}=a+b\sqrt{2}$$

10-3\sqrt{2} ga \sqrt{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\frac{10\sqrt{2}-3\left(\sqrt{2}\right)^{2}}{2}=a+b\sqrt{2}$$

\sqrt{2} kvadrati – 2.

$$\frac{10\sqrt{2}-3\times 2}{2}=a+b\sqrt{2}$$

-6 hosil qilish uchun -3 va 2 ni ko'paytirish.

$$\frac{10\sqrt{2}-6}{2}=a+b\sqrt{2}$$

5\sqrt{2}-3 natijani olish uchun 10\sqrt{2}-6 ning har bir ifodasini 2 ga bo‘ling.

$$5\sqrt{2}-3=a+b\sqrt{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$a+b\sqrt{2}=5\sqrt{2}-3$$

Ikkala tarafdan a ni ayirish.

$$b\sqrt{2}=5\sqrt{2}-3-a$$

Tenglama standart shaklda.

$$\sqrt{2}b=-a+5\sqrt{2}-3$$

Ikki tarafini \sqrt{2} ga bo‘ling.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+5\sqrt{2}-3}{\sqrt{2}}$$

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

$$b=\frac{-a+5\sqrt{2}-3}{\sqrt{2}}$$

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

$$b=\frac{\sqrt{2}\left(-a+5\sqrt{2}-3\right)}{2}$$

a uchun yechish

$$a=-\sqrt{2}b+5\sqrt{2}-3$$