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\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=a+b\sqrt{3}
\frac{2+\sqrt{3}}{2-\sqrt{3}} maxrajini 2+\sqrt{3} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.
\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{2^{2}-\left(\sqrt{3}\right)^{2}}=a+b\sqrt{3}
Hisoblang: \left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{4-3}=a+b\sqrt{3}
2 kvadratini chiqarish. \sqrt{3} kvadratini chiqarish.
\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{1}=a+b\sqrt{3}
1 olish uchun 4 dan 3 ni ayirish.
\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)=a+b\sqrt{3}
Har qanday son birga bo‘linganda, natija o‘zi chiqadi.
\left(2+\sqrt{3}\right)^{2}=a+b\sqrt{3}
\left(2+\sqrt{3}\right)^{2} hosil qilish uchun 2+\sqrt{3} va 2+\sqrt{3} ni ko'paytirish.
4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}=a+b\sqrt{3}
\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2+\sqrt{3}\right)^{2} kengaytirilishi uchun ishlating.
4+4\sqrt{3}+3=a+b\sqrt{3}
\sqrt{3} kvadrati – 3.
7+4\sqrt{3}=a+b\sqrt{3}
7 olish uchun 4 va 3'ni qo'shing.
a+b\sqrt{3}=7+4\sqrt{3}
Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.
b\sqrt{3}=7+4\sqrt{3}-a
Ikkala tarafdan a ni ayirish.
\sqrt{3}b=-a+4\sqrt{3}+7
Tenglama standart shaklda.
\frac{\sqrt{3}b}{\sqrt{3}}=\frac{-a+4\sqrt{3}+7}{\sqrt{3}}
Ikki tarafini \sqrt{3} ga bo‘ling.
b=\frac{-a+4\sqrt{3}+7}{\sqrt{3}}
\sqrt{3} ga bo'lish \sqrt{3} ga ko'paytirishni bekor qiladi.
b=\frac{\sqrt{3}\left(-a+4\sqrt{3}+7\right)}{3}
4\sqrt{3}-a+7 ni \sqrt{3} ga bo'lish.