Question

Stiind ca a-b=1, calculati $a^{2} + b^{2}$ -2ab+2a-2b+1

Answer 1

a²+b²-2ab+2a-2b+1=
=(a-b)²+2(a-b)+1=(a-b+1)²=(1+1)²=4
sau
a²+b²-2ab+2a-2b+1=
=(a-b)²+2(a-b)+1=1+2+1=4

Answer 2

[tex]a=1-b \\ (1-b)^{2} + b^{2} -b(1-b)+2(1-b)-2b+1 \\ 1-2b+ b^{2} + b^{2} -b+ b^{2} +2-2b-2b+1 \\ 2b^{2} -6b+4=0 \\ delta=36-32=4 \\ x_{1} = \frac{6+2}{4} =2 \\ x_{2} = \frac{6-2}{4} =1 \\ daca b=2 =\ \textgreater \ a=3 \\ daca b=1=\ \textgreater \ a=0[/tex]