Question

Mình đang cần gấp!!!

Chứng minh : a) \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

b) \(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
 

Answer 1

a) \(\Leftrightarrow\left(\sqrt{3+\sqrt{5}}\right)^2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

\(\Leftrightarrow\sqrt{3+\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)\cdot\left(3+\sqrt{5}\right)}=8\)

\(\Leftrightarrow\sqrt{\frac{6+2\sqrt{5}}{2}}\cdot\left(\sqrt{5}\sqrt{2}-\sqrt{2}\right)\sqrt{3^2-5}=8\).

\(\Leftrightarrow\sqrt{\frac{5+2\sqrt{5}+1}{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{4}=8\)

\(\Leftrightarrow\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\cdot2=8\)

\(\Leftrightarrow\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\Leftrightarrow\left(\sqrt{5}\right)^2-1=4\Leftrightarrow5-1=4\)Đúng -ĐPCM.

Answer 2

Cậu giải dùm mình câu b luôn nhé! cảm ơn c! :)))))))

Answer 3

b) Rõ ràng: \(\sqrt{\sqrt{2}+1}>\sqrt{\sqrt{2}-1}\)

Bình phương 2 vế ta được:

\(\sqrt{2}+1-2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{2}-1=2\left(\sqrt{2}-1\right)\)

\(\Leftrightarrow2\sqrt{2}-2\sqrt{\left(\sqrt{2}\right)^2-1}=2\sqrt{2}-2\)

\(\Leftrightarrow2\sqrt{2}-2=2\sqrt{2}-2\)Hiển nhiên đúng. ĐPCM