[1+1/sqrt{2}+1/2][1-1/sqrt{2}+1/2] eching

$$[ 1 + \frac { 1 } { \sqrt { 2 } } + \frac { 1 } { 2 } ] [ 1 - \frac { 1 } { \sqrt { 2 } } + \frac { 1 } { 2 } ]$$

Baholash

$$\frac{7}{4}=1,75$$

Yechim qadamlari

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

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

\sqrt{2} kvadrati – 2.

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

1 ni \frac{2}{2} kasrga o‘giring.

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

\frac{2}{2} va \frac{1}{2} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

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

3 olish uchun 2 va 1'ni qo'shing.

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

\frac{3}{2} va \frac{\sqrt{2}}{2} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

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

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

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

\sqrt{2} kvadrati – 2.

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

1 ni \frac{2}{2} kasrga o‘giring.

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

\frac{2}{2} va \frac{1}{2} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

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

3 olish uchun 2 va 1'ni qo'shing.

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

\frac{3}{2} va \frac{\sqrt{2}}{2} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

$$\frac{3+\sqrt{2}}{2}\times \frac{3+\sqrt{2}}{2}$$

\left(\frac{3+\sqrt{2}}{2}\right)^{2} hosil qilish uchun \frac{3+\sqrt{2}}{2} va \frac{3+\sqrt{2}}{2} ni ko'paytirish.

$$\left(\frac{3+\sqrt{2}}{2}\right)^{2}$$

\frac{3+\sqrt{2}}{2}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{\left(3+\sqrt{2}\right)^{2}}{2^{2}}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3+\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{2^{2}}$$

\sqrt{2} kvadrati – 2.

$$\frac{9+6\sqrt{2}+2}{2^{2}}$$

11 olish uchun 9 va 2'ni qo'shing.

$$\frac{11+6\sqrt{2}}{2^{2}}$$

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

$$\frac{11+6\sqrt{2}}{4}$$

Omil

$$\frac{7}{2 ^ {2}} = 1\frac{3}{4} = 1,75$$