Bài 22:
a: \(A=x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\)
Khi x=103 thì \(A=\left(103-3\right)^2+1=100^2+1=10000+1=10001\)
b: \(B=x^2+4x+1\)
\(=x^2+4x+4-3\)
\(=\left(x+2\right)^2-3\)
\(=\left(98+2\right)^2-3=100^2-3=10000-3=9997\)
Bài 21:
\(\left(10^2+25\right)^2-\left(10^2-25\right)^2=10^{n}\)
=>\(10^{n}=\left(10^2+25-10^2+25\right)\left(10^2+25+10^2-25\right)\)
=>\(10^{n}=50\cdot200=10000=10^4\)
=>n=4
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