Tìm a, b, c biết: a^2 - 2a + b^2 + 4b + 4c^2 - 4c + 6 = 0
Giải thích
Ta có:
a2−2a+b2+4b+4c2−4c+6=0a2−2a+1+b2+4b+4+4c2−4c+1=0a−12+b+22+2c−12=0Do a−12≥0,b+22≥0;2c−12≥0⇒a−1=0b+2=02c−1=0⇔a=1b=−2c=12
Vậy a=1;b=−2;c=12
Ta có:
a2−2a+b2+4b+4c2−4c+6=0a2−2a+1+b2+4b+4+4c2−4c+1=0a−12+b+22+2c−12=0Do a−12≥0,b+22≥0;2c−12≥0⇒a−1=0b+2=02c−1=0⇔a=1b=−2c=12
Vậy a=1;b=−2;c=12