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a, \(=\frac{8\left(x+2y\right)^5}{2\left(x+2y\right)}=8\left(x+2y\right)^4\)
b,\(=\left(\frac{3}{4}\right)^{45}:\left(\left(\frac{3}{4}\right)^2\right)^{10}\)
\(=\left(\frac{3}{4}\right)^{45}:\left(\frac{3}{4}\right)^{2.10}=\left(\frac{3}{4}\right)^{25}\)
2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
chúc bn hok tốt @_@
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)
a) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)
b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\\ =6ab^2\)
\(9^8.2^8-\left(184-1\right)\left(184+1\right)=18^8-\left(184^2-1\right)=18^8-184^2+1\)