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19 tháng 11 2016
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^{64}-1\right)+11=3^{64}+10\)
19 tháng 11 2016
A = 8.(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (38 - 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (316 - 1)(316 + 1)(332 + 1) + 1
A = (332 - 1)(332 + 1) + 1
A = 364 - 1 + 1
A = 364
14 tháng 6 2018
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
TT
1
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
NH
0
NQ
1
11 tháng 7 2017
nhan ca tu va mau voi\(\sqrt{2}\) ta dc
\(\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}\)(dkx>=2)
=\(\frac{\left|\sqrt{2x-4}-2\right|}{2}\)
\(2\sqrt{18}+3\sqrt{32}-4\sqrt{8}\)
\(=6\sqrt{2}+12\sqrt{2}-8\sqrt{2}\)
\(=10\sqrt{2}\)
\(2\sqrt{18}+3\sqrt{32}-4\sqrt{8}\)
\(=2.3\sqrt{2}+3.4\sqrt{2}-4.2\sqrt{2}\)(\(3\sqrt{32}=3.4\sqrt{2}\)vì \(\sqrt{32}\)= \(2\sqrt{8}\)mà \(\sqrt{8}=2\sqrt{2}\))
\(=6\sqrt{2}+12\sqrt{2}-8\sqrt{2}\)
\(=\sqrt{2}\left(6+12-8\right)\)
\(=10\sqrt{2}\)