rut gon bieu thuc sau : 4^2.25^2+32.125/2^3.5^2
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WT
1
4 tháng 8 2017
\(B=\frac{4^2.25^2+32.125}{2^3.5^2}=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}\)
\(=\frac{2^4.5^3\left(5+2\right)}{2^3.5^2}=\frac{2^3.5^2.2.5.7}{2^3.5^2}=2.5.7=70\)
VN
0
4 tháng 10 2021
\(\dfrac{4^2.25^2+32.125}{2^3.5^2}=\dfrac{2^4.5^4+2^5.5^3}{2^3.5^2}=\dfrac{2^3.5^2\left(2.5^2+2^2.5\right)}{2^3.5^2}=2.5^2+2^2.5=50+20=70\)
25 tháng 7 2015
\(\frac{4^{72}.5^{24}}{2^{16}}=\frac{2^{144}.5^{24}}{2^{16}}=2^{128}.5^{24}\)
TD
0
HT
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: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
5 tháng 10 2019
\(3\left(2^2+1\right).\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right).\left(2^8+1\right)\left(2^{16}+1\right)....\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right).\left(2^{64}+1\right)+1\)
\(=2^{64}-1+1=2^{64}\)
Vậy : \(3\left(2^2+1\right).\left(2^4+1\right)...\left(2^{64}+1\right)+1=2^{64}\)