Bafi1:
\(\left(3x+4\right)\left(9x^2-12x+16\right)=65\)
<=>\(27x^3+64=65\)
=>\(27x^3=1\)
=>\(x^3=\dfrac{1}{27}\)
=>\(x=\dfrac{1}{3}\)
Vậy...
Bafi2:
\(M=\left(x+y-1\right)^3-\left(x+y+1\right)^3+6\left(x+y\right)^2\)
\(=-2-6x^2-12xy-6y^2+6\left(x^2+2xy+y^2\right)\)
\(=-2\)
Vậy...(đpcm)