(sqrt{5}-sqrt{3})(sqrt{5}+sqrt{3})-(sqrt{6}+sqrt{2})^2 eching

$$( \sqrt { 5 } - \sqrt { 3 } ) ( \sqrt { 5 } + \sqrt { 3 } ) - ( \sqrt { 6 } + \sqrt { 2 } ) ^ { 2 }$$

Baholash

$$-4\sqrt{3}-6\approx -12,92820323$$

Yechim qadamlari

Hisoblang: \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}-\left(\sqrt{6}+\sqrt{2}\right)^{2}$$

\sqrt{5} kvadrati – 5.

$$5-\left(\sqrt{3}\right)^{2}-\left(\sqrt{6}+\sqrt{2}\right)^{2}$$

\sqrt{3} kvadrati – 3.

$$5-3-\left(\sqrt{6}+\sqrt{2}\right)^{2}$$

2 olish uchun 5 dan 3 ni ayirish.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(\sqrt{6}+\sqrt{2}\right)^{2} kengaytirilishi uchun ishlating.

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

\sqrt{6} kvadrati – 6.

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

Faktor: 6=2\times 3. \sqrt{2\times 3} koʻpaytmasining kvadrat ildizini \sqrt{2}\sqrt{3} kvadrat ildizlarining koʻpaytmasi sifatida qayta yozing.

$$2-\left(6+2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

2 hosil qilish uchun \sqrt{2} va \sqrt{2} ni ko'paytirish.

$$2-\left(6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)$$

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

$$2-\left(6+4\sqrt{3}+\left(\sqrt{2}\right)^{2}\right)$$

\sqrt{2} kvadrati – 2.

$$2-\left(6+4\sqrt{3}+2\right)$$

8 olish uchun 6 va 2'ni qo'shing.

$$2-\left(8+4\sqrt{3}\right)$$

8+4\sqrt{3} teskarisini topish uchun har birining teskarisini toping.

$$2-8-4\sqrt{3}$$

-6 olish uchun 2 dan 8 ni ayirish.

$$-6-4\sqrt{3}$$