Bài 1:
\(x^2+5y^2+2x-4xy-10y+2014\)
\(=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)+2004\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+2004\)
Do \(\left(x-2y+1\right)^2\ge0;\left(y-3\right)^2\ge0\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2+2004>0\)
=> đpcm
Bài 2:
\(A=3.\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)
...
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=\left(2^{128}-1\right)+1=2^{128}\)
Vậy \(A=2^{128}\)
B1:
\(x^2+5y^2+2x-4xy-10y+2014\)
= \(\left(x^2+2x-4xy+1-4y+4y^2\right)+y^2-6y+9+2004\)
= \(\left[x^2+2x\left(1-2y\right)+\left(1-2y\right)^2\right]+\left(y-3\right)^2+2004\)
= \(\left(x-1+2y\right)^2+\left(y-3\right)^2+2004>0\) với mọi x
\(\Rightarrow\) Đpcm
B2:
\(A=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^{16}-1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^{32}-1\right).....\left(2^{64}+1\right)+1\)
\(A=\left(2^{64}-1\right).\left(2^{64}+1\right)+1\)
\(A=2^{128}-1+1=2^{128}\)
