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(sqrt(2) + sqrt(3))2 = (a+b)2

Mathematics

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Answer

5+265 + 2\sqrt{6}

Step-by-step solution

4 steps

  1. दिए गए व्यंजक और बीजगणितीय सर्वसमिका को पहचानें।

    व्यंजक (2+3)2(\sqrt{2} + \sqrt{3})^2 है।
    हमें इसे (a+b)2(a+b)^2 के रूप में विस्तारित करना है।
    बीजगणितीय सर्वसमिका है: (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

  2. सर्वसमिका में aa और bb के मानों को प्रतिस्थापित करें।

    यहाँ, a=2a = \sqrt{2} और b=3b = \sqrt{3} है।

    (2+3)2=(2)2+2(2)(3)+(3)2(\sqrt{2} + \sqrt{3})^2 = (\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2

  3. प्रत्येक पद को सरल करें।

    (2)2=2(\sqrt{2})^2 = 2 (3)2=3(\sqrt{3})^2 = 3 2(2)(3)=22×3=262(\sqrt{2})(\sqrt{3}) = 2\sqrt{2 \times 3} = 2\sqrt{6}

  4. सरलीकृत पदों को एक साथ जोड़ें।

    2+26+32 + 2\sqrt{6} + 3 (2+3)+26(2 + 3) + 2\sqrt{6} 5+265 + 2\sqrt{6}
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