Question

cho x³+27y³=1-9xy(x+3y)

tính M=x+3y

Answer 1

\(x^3+27y^3=1-9xy\left(x+3y\right)\)

<=>  \(x^3+27y^3+9xy\left(x+3y\right)=1\)

<=>  \(\left(x+3y\right)^3=1\)

<=>  \(x+3y=1\)

Vậy  \(M=1\)

Answer 2

\(x^3+27x^3=1-9xy\left(x+3y\right)\))

\(=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=1-9xy\left(x+3y\right)\)

=\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)-1+9xy\left(x+3y\right)=0\)

=\(\left(x+3y\right)\left(x^2-3xy+9y^2+9xy\right)-1=0\)

=\(\left(x+3y\right)\left(x^2+6xy+9y^2\right)-1=0\)

=\(\left(x+3y\right)\left(x+3y\right)^2-1=0\)

=\(\left(x+3y\right)\left(x+3y\right)^2=1\)

\(\Rightarrow x+3y=\left(x+3y\right)^2=1\)

\(\Rightarrow x+3y=1\)