\(a^2-2a+b^2+4b+4c^2-4c+6=0\)'
\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
b tự làm nốt nhé~
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)
\(M=x^3+3^3-x^3-54+x\)
\(M=x+27-54\)
\(M=x+27-54\)
\(M=7-27\)
\(M=-20\)
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)
\(M=x^3+3^3-x^3-54+x\)
\(M=27-54+x\)
\(M=-27+x\)
thay x =7 vào M ta có :
\(M=-27+7=-20\)
a2-2a+b2+4b+4c2-4c+6=0
(a2-2a+1)+(b2+4b+4)+(4c2-4c+1)=0
(a-1)2+(b+2)2+(2c-1)2=0
Còn lại cậu tự làm nhé mik bận!