-frac{(sqrt{2}-1)^2}{4sqrt{2}}+frac{(sqrt{5}+sqrt{3})^2}{sqrt{15}}+frac{(sqrt{2}+1)^2}{4sqrt{2}}-frac{(sqrt{5}-sqrt{3})^2}{sqrt{15}} সমাধান কৰক

$$- \frac { ( \sqrt { 2 } - 1 ) ^ { 2 } } { 4 \sqrt { 2 } } + \frac { ( \sqrt { 5 } + \sqrt { 3 } ) ^ { 2 } } { \sqrt { 15 } } + \frac { ( \sqrt { 2 } + 1 ) ^ { 2 } } { 4 \sqrt { 2 } } - \frac { ( \sqrt { 5 } - \sqrt { 3 } ) ^ { 2 } } { \sqrt { 15 } }$$

মূল্যায়ন

$$5$$

সমাধানৰ স্তৰবোৰ সজাওক

\left(\sqrt{2}-1\right)^{2} বিস্তাৰ কৰিবলৈ দ্বিপদীয় উপপাদ্য \left(a-b\right)^{2}=a^{2}-2ab+b^{2} ব্যৱহাৰ কৰক৷

$$-\frac{\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1}{4\sqrt{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{2}ৰ বৰ্গমূল হৈছে 2৷

$$-\frac{2-2\sqrt{2}+1}{4\sqrt{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

3 লাভ কৰিবৰ বাবে 2 আৰু 1 যোগ কৰক৷

$$-\frac{3-2\sqrt{2}}{4\sqrt{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

হৰ আৰু লৱক \sqrt{2}ৰে পূৰণ কৰি \frac{3-2\sqrt{2}}{4\sqrt{2}}ৰ হৰৰ মূল উলিয়াওক।

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{2}ৰ বৰ্গমূল হৈছে 2৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{4\times 2}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

8 লাভ কৰিবৰ বাবে 4 আৰু 2 পুৰণ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\left(\sqrt{5}+\sqrt{3}\right)^{2} বিস্তাৰ কৰিবলৈ দ্বিপদীয় উপপাদ্য \left(a+b\right)^{2}=a^{2}+2ab+b^{2} ব্যৱহাৰ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{5}ৰ বৰ্গমূল হৈছে 5৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{5} আৰু \sqrt{3}ক পূৰণ কৰিবলৈ, সংখ্যাবোৰ বৰ্গমূলৰ তলত পূৰণ কৰক।

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{3}ৰ বৰ্গমূল হৈছে 3৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{5+2\sqrt{15}+3}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

8 লাভ কৰিবৰ বাবে 5 আৰু 3 যোগ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{8+2\sqrt{15}}{\sqrt{15}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

হৰ আৰু লৱক \sqrt{15}ৰে পূৰণ কৰি \frac{8+2\sqrt{15}}{\sqrt{15}}ৰ হৰৰ মূল উলিয়াওক।

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{\left(\sqrt{15}\right)^{2}}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{15}ৰ বৰ্গমূল হৈছে 15৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(\sqrt{2}+1\right)^{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\left(\sqrt{2}+1\right)^{2} বিস্তাৰ কৰিবলৈ দ্বিপদীয় উপপাদ্য \left(a+b\right)^{2}=a^{2}+2ab+b^{2} ব্যৱহাৰ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{2}ৰ বৰ্গমূল হৈছে 2৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{2+2\sqrt{2}+1}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

3 লাভ কৰিবৰ বাবে 2 আৰু 1 যোগ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{3+2\sqrt{2}}{4\sqrt{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

হৰ আৰু লৱক \sqrt{2}ৰে পূৰণ কৰি \frac{3+2\sqrt{2}}{4\sqrt{2}}ৰ হৰৰ মূল উলিয়াওক।

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{4\left(\sqrt{2}\right)^{2}}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{2}ৰ বৰ্গমূল হৈছে 2৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{4\times 2}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

8 লাভ কৰিবৰ বাবে 4 আৰু 2 পুৰণ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\left(\sqrt{5}-\sqrt{3}\right)^{2} বিস্তাৰ কৰিবলৈ দ্বিপদীয় উপপাদ্য \left(a-b\right)^{2}=a^{2}-2ab+b^{2} ব্যৱহাৰ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{5}ৰ বৰ্গমূল হৈছে 5৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{5} আৰু \sqrt{3}ক পূৰণ কৰিবলৈ, সংখ্যাবোৰ বৰ্গমূলৰ তলত পূৰণ কৰক।

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{\sqrt{15}}$$

\sqrt{3}ৰ বৰ্গমূল হৈছে 3৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{5-2\sqrt{15}+3}{\sqrt{15}}$$

8 লাভ কৰিবৰ বাবে 5 আৰু 3 যোগ কৰক৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{8-2\sqrt{15}}{\sqrt{15}}$$

হৰ আৰু লৱক \sqrt{15}ৰে পূৰণ কৰি \frac{8-2\sqrt{15}}{\sqrt{15}}ৰ হৰৰ মূল উলিয়াওক।

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{\left(\sqrt{15}\right)^{2}}$$

\sqrt{15}ৰ বৰ্গমূল হৈছে 15৷

$$-\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8}+\frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

এক্সপ্ৰেশ্বন যোগ বা বিয়োগ কৰিবলৈ, সিহঁতৰ হৰ একে কৰিবলৈ বিস্তাৰ কৰক৷ 8 আৰু 15ৰ সাধাৰণ গুণফল হৈছে 120৷ -\frac{\left(3-2\sqrt{2}\right)\sqrt{2}}{8} বাৰ \frac{15}{15} পুৰণ কৰক৷ \frac{\left(8+2\sqrt{15}\right)\sqrt{15}}{15} বাৰ \frac{8}{8} পুৰণ কৰক৷

$$-\frac{15\left(3-2\sqrt{2}\right)\sqrt{2}}{120}+\frac{8\left(8+2\sqrt{15}\right)\sqrt{15}}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

যিহেতু -\frac{15\left(3-2\sqrt{2}\right)\sqrt{2}}{120} আৰু \frac{8\left(8+2\sqrt{15}\right)\sqrt{15}}{120}ৰ একে ডেনোমিনেটৰ আছে, গতিকে সিহঁতক সিহঁতৰ নিউমেৰেটৰ যোগ কৰি যোগ কৰক৷

$$\frac{-15\left(3-2\sqrt{2}\right)\sqrt{2}+8\left(8+2\sqrt{15}\right)\sqrt{15}}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

-15\left(3-2\sqrt{2}\right)\sqrt{2}+8\left(8+2\sqrt{15}\right)\sqrt{15}ত গুণনিয়ক কৰক৷

$$\frac{-45\sqrt{2}+60+64\sqrt{15}+240}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

-45\sqrt{2}+60+64\sqrt{15}+240ত গণনা কৰক৷

$$\frac{-45\sqrt{2}+300+64\sqrt{15}}{120}+\frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

এক্সপ্ৰেশ্বন যোগ বা বিয়োগ কৰিবলৈ, সিহঁতৰ হৰ একে কৰিবলৈ বিস্তাৰ কৰক৷ 120 আৰু 8ৰ সাধাৰণ গুণফল হৈছে 120৷ \frac{\left(3+2\sqrt{2}\right)\sqrt{2}}{8} বাৰ \frac{15}{15} পুৰণ কৰক৷

$$\frac{-45\sqrt{2}+300+64\sqrt{15}}{120}+\frac{15\left(3+2\sqrt{2}\right)\sqrt{2}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

যিহেতু \frac{-45\sqrt{2}+300+64\sqrt{15}}{120} আৰু \frac{15\left(3+2\sqrt{2}\right)\sqrt{2}}{120}ৰ একে ডেনোমিনেটৰ আছে, গতিকে সিহঁতক সিহঁতৰ নিউমেৰেটৰ যোগ কৰি যোগ কৰক৷

$$\frac{-45\sqrt{2}+300+64\sqrt{15}+15\left(3+2\sqrt{2}\right)\sqrt{2}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

-45\sqrt{2}+300+64\sqrt{15}+15\left(3+2\sqrt{2}\right)\sqrt{2}ত গুণনিয়ক কৰক৷

$$\frac{-45\sqrt{2}+300+64\sqrt{15}+45\sqrt{2}+60}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

-45\sqrt{2}+300+64\sqrt{15}+45\sqrt{2}+60ত গণনা কৰক৷

$$\frac{360+64\sqrt{15}}{120}-\frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15}$$

এক্সপ্ৰেশ্বন যোগ বা বিয়োগ কৰিবলৈ, সিহঁতৰ হৰ একে কৰিবলৈ বিস্তাৰ কৰক৷ 120 আৰু 15ৰ সাধাৰণ গুণফল হৈছে 120৷ \frac{\left(8-2\sqrt{15}\right)\sqrt{15}}{15} বাৰ \frac{8}{8} পুৰণ কৰক৷

$$\frac{360+64\sqrt{15}}{120}-\frac{8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}$$

যিহেতু \frac{360+64\sqrt{15}}{120} আৰু \frac{8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}ৰ একে ডেনোমিনেটৰ আছে, গতিকে সিহঁতক সিহঁতৰ নিউমেৰেটৰ বিয়োগ কৰি বিয়োগ কৰক৷

$$\frac{360+64\sqrt{15}-8\left(8-2\sqrt{15}\right)\sqrt{15}}{120}$$

360+64\sqrt{15}-8\left(8-2\sqrt{15}\right)\sqrt{15}ত গুণনিয়ক কৰক৷

$$\frac{360+64\sqrt{15}-64\sqrt{15}+240}{120}$$

360+64\sqrt{15}-64\sqrt{15}+240ত গণনা কৰক৷

$$\frac{600}{120}$$

5 লাভ কৰিবলৈ 120ৰ দ্বাৰা 600 হৰণ কৰক৷

$$5$$

কাৰক

$$5$$