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\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{1}{2+\sqrt{3}}+\frac{\sqrt{8}}{\sqrt{2}}
\frac{1}{2-\sqrt{3}} maxrajini 2+\sqrt{3} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.
\frac{2+\sqrt{3}}{2^{2}-\left(\sqrt{3}\right)^{2}}+\frac{1}{2+\sqrt{3}}+\frac{\sqrt{8}}{\sqrt{2}}
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{2+\sqrt{3}}{4-3}+\frac{1}{2+\sqrt{3}}+\frac{\sqrt{8}}{\sqrt{2}}
2 kvadratini chiqarish. \sqrt{3} kvadratini chiqarish.
\frac{2+\sqrt{3}}{1}+\frac{1}{2+\sqrt{3}}+\frac{\sqrt{8}}{\sqrt{2}}
1 olish uchun 4 dan 3 ni ayirish.
2+\sqrt{3}+\frac{1}{2+\sqrt{3}}+\frac{\sqrt{8}}{\sqrt{2}}
Har qanday son birga bo‘linganda, natija o‘zi chiqadi.
2+\sqrt{3}+\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{8}}{\sqrt{2}}
\frac{1}{2+\sqrt{3}} maxrajini 2-\sqrt{3} orqali surat va maxrajini koʻpaytirish orqali ratsionallashtiring.
2+\sqrt{3}+\frac{2-\sqrt{3}}{2^{2}-\left(\sqrt{3}\right)^{2}}+\frac{\sqrt{8}}{\sqrt{2}}
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}.
2+\sqrt{3}+\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{8}}{\sqrt{2}}
2 kvadratini chiqarish. \sqrt{3} kvadratini chiqarish.
2+\sqrt{3}+\frac{2-\sqrt{3}}{1}+\frac{\sqrt{8}}{\sqrt{2}}
1 olish uchun 4 dan 3 ni ayirish.
2+\sqrt{3}+2-\sqrt{3}+\frac{\sqrt{8}}{\sqrt{2}}
Har qanday son birga bo‘linganda, natija o‘zi chiqadi.
4+\sqrt{3}-\sqrt{3}+\frac{\sqrt{8}}{\sqrt{2}}
4 olish uchun 2 va 2'ni qo'shing.
4+\frac{\sqrt{8}}{\sqrt{2}}
0 ni olish uchun \sqrt{3} va -\sqrt{3} ni birlashtirish.
4+\sqrt{4}
\frac{\sqrt{8}}{\sqrt{2}} kvadrat ildizlar boʻlinmasini \sqrt{\frac{8}{2}} boʻlinmasining kvadrat ildizi sifatida qayta yozing va boʻlishni amalga oshiring.
4+2
4 ning kvadrat ildizini hisoblab, 2 natijaga ega bo‘ling.
6
6 olish uchun 4 va 2'ni qo'shing.