Примените формулу квадрата суммы:
- \((a+1)^2=a^2+2\cdot{a}\cdot1+1^2=a^2+2a+1;\)
- \((2a+1)^2=(2a)^2+2\cdot2a\cdot1+1^2=4a^2+4a+1\)
- \({\largeа)}\ (a+2)^2=a^2+\ \)\(\ +2^2=\ \)
- \({\largeб)}\ (a+3)^2=a^2+\ \)
- \({\largeв)}\ (a+4)^2=\ \)
- \({\largeг)}\ (a+5)^2=\ \)
- \({\largeд)}\ (3a+1)^2=\ \)
- \({\largeе)}\ (2a+3)^2=\ \)
- \({\largeж)}\ (3a+0{,}5)^2=\ \)