B=frac{sqrt{2}-sqrt{7}}{5-sqrt{8}} eching

$$B = \frac { \sqrt { 2 } - \sqrt { 7 } } { 5 - \sqrt { 8 } }$$

B uchun yechish

$$B=\frac{5\sqrt{2}+4-2\sqrt{14}-5\sqrt{7}}{17}\approx -0,567117854$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Faktor: 8=2^{2}\times 2. \sqrt{2^{2}\times 2} koʻpaytmasining kvadrat ildizini \sqrt{2^{2}}\sqrt{2} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing. 2^{2} ning kvadrat ildizini chiqarish.

$$B=\frac{\sqrt{2}-\sqrt{7}}{5-2\sqrt{2}}$$

\frac{\sqrt{2}-\sqrt{7}}{5-2\sqrt{2}} maxrajini 5+2\sqrt{2} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{\left(5-2\sqrt{2}\right)\left(5+2\sqrt{2}\right)}$$

Hisoblang: \left(5-2\sqrt{2}\right)\left(5+2\sqrt{2}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{5^{2}-\left(-2\sqrt{2}\right)^{2}}$$

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{25-\left(-2\sqrt{2}\right)^{2}}$$

\left(-2\sqrt{2}\right)^{2} ni kengaytirish.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{25-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}$$

2 daraja ko‘rsatkichini -2 ga hisoblang va 4 ni qiymatni oling.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{25-4\left(\sqrt{2}\right)^{2}}$$

\sqrt{2} kvadrati – 2.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{25-4\times 2}$$

8 hosil qilish uchun 4 va 2 ni ko'paytirish.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{25-8}$$

17 olish uchun 25 dan 8 ni ayirish.

$$B=\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(5+2\sqrt{2}\right)}{17}$$

\sqrt{2}-\sqrt{7} ifodaning har bir elementini 5+2\sqrt{2} ifodaning har bir elementiga ko‘paytirish orqali taqsimot qonuni xususiyatlarini qo‘llash mumkin.

$$B=\frac{5\sqrt{2}+2\left(\sqrt{2}\right)^{2}-5\sqrt{7}-2\sqrt{7}\sqrt{2}}{17}$$

\sqrt{2} kvadrati – 2.

$$B=\frac{5\sqrt{2}+2\times 2-5\sqrt{7}-2\sqrt{7}\sqrt{2}}{17}$$

4 hosil qilish uchun 2 va 2 ni ko'paytirish.

$$B=\frac{5\sqrt{2}+4-5\sqrt{7}-2\sqrt{7}\sqrt{2}}{17}$$

\sqrt{7} va \sqrt{2} ni koʻpaytirish uchun kvadrat ildiz ichidagi sonlarni koʻpaytiring.

$$B=\frac{5\sqrt{2}+4-5\sqrt{7}-2\sqrt{14}}{17}$$

\frac{5}{17}\sqrt{2}+\frac{4}{17}-\frac{5}{17}\sqrt{7}-\frac{2}{17}\sqrt{14} natijani olish uchun 5\sqrt{2}+4-5\sqrt{7}-2\sqrt{14} ning har bir ifodasini 17 ga bo‘ling.

$$B=\frac{5}{17}\sqrt{2}+\frac{4}{17}-\frac{5}{17}\sqrt{7}-\frac{2}{17}\sqrt{14}$$

B'ni tayinlash (complex solution)

$$B≔\frac{\left(\sqrt{2}-\sqrt{7}\right)\left(2\sqrt{2}+5\right)}{17}$$

B'ni tayinlash

$$B≔\frac{5\sqrt{2}+4-2\sqrt{14}-5\sqrt{7}}{17}$$