Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt{1}=1\)
\(\sqrt{1+2+1}=2\)
\(\sqrt{1+2+3+2+1}=3\)
b) \(\sqrt{1+2+3+4+3+2+1}=4\)
\(\sqrt{1+2+3+4+5+4+3+2+1}=5\)
\(\sqrt{1+2+3+4+5+6+5+4+3+2+1}=6\)
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
.\(\sqrt{1}=1\)
\(\sqrt{1+2+1}=\sqrt{4}=2\)
\(\sqrt{1+2+3+2+1}=\sqrt{9}=3\)
b,
\(\sqrt{1+2+3+4+3+2+1}=\sqrt{16}=4\)
\(\sqrt{1+2+3+4+5+4+3+2+1}=\sqrt{25}=5\)
\(\sqrt{1+2+3+4+5+6+5+4+3+2+1}=\sqrt{36}=6\)
*Nhận xét:
+\(\sqrt{1+...+10+...1}=10\)
+\(\sqrt{1+2+...+100+1}=100\)
+\(\sqrt{1+2+...n+...1}=n;n\in N\)*
a) \(-\frac{48}{625}\)
b) không có giá trị
c) 1:125
d) \(-\frac{1053}{500}\)
e) 1
g) -13
a) \(-\dfrac{48}{625}\)
b) \(\varnothing\)
c) \(\dfrac{1}{125}\)
d) \(-\dfrac{1053}{500}\)
e) 1
f) -13
a) \(\sqrt{121}=11\)
\(\sqrt{12321}=111\)
\(\sqrt{1234321}=1111\)
b) \(\sqrt{123454321}=11111\)
\(\sqrt{12345654321}=111111\)
\(\sqrt{1234567654321}=1111111\)