pn lấy đề ở đâu vậy ?

Ở lớp học thêm c ạ

A= ( \(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\))

= (1+1,4+1,7)+(4,4+6,3+7,7)

= 4,1+18,4

=22,5

Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)

Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.

a) A<B

A<B

k cho minh nhe

a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)

\(\sqrt{6}< \sqrt{6,25}=2,5\); 

\(\sqrt{12}< \sqrt{12,25}=3,5\); 

\(\sqrt{20}< \sqrt{20,25}=4,5\)

=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)

Vậy P < 12

Answer:

ý a, tham khảo bài làm của @xyzquynhdi

\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)

\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)

a/ \(\sqrt{2}+\sqrt{6}\)

b/ Sửa đề:

\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)

c/ \(1+\sqrt{2}+\sqrt{5}\)

giải rõ ra hộ mình với

a/ \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)

\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}\)

\(=2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}\)

\(=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=2\sqrt{3+\left(\sqrt{3}-1\right)}\)

\(=\sqrt{2}\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{2}\sqrt{3+2\sqrt{3}+1}\)

\(=\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{2}\left(\sqrt{3}+1\right)\)

\(=\sqrt{2}+\sqrt{6}\)

Bạn tham khảo tại link sau:

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