CMR x+y+z+t=0 thì x^3+y^3+z^3+t^3= 3(xyt)(z+t) 30/09/2021 Bởi Kylie CMR x+y+z+t=0 thì x^3+y^3+z^3+t^3= 3(xyt)(z+t)
\(x+y+z+t=0\Rightarrow t=-\left(x+y+z\right)\) Ta có: \(VT=x^3+y^3+z^3+t^3=x^3+y^3+z^3-\left(x+y+z\right)^3\) \(=x^3+y^3+z^3-\left[x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]\) \(=-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\) \(VP=3\left[xy+z\left(x+y+z\right)\right]\left(z-x-y-z\right)=3\left(xy+yz+zx+z^2\right)\left(-x-y\right)\) \(=-3\left(y+z\right)\left(z+x\right)\left(x+y\right)\) Do VT = VP nên ta có đpcm. Bình luận
\(x+y+z+t=0\Rightarrow t=-\left(x+y+z\right)\)
Ta có:
\(VT=x^3+y^3+z^3+t^3=x^3+y^3+z^3-\left(x+y+z\right)^3\)
\(=x^3+y^3+z^3-\left[x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]\)
\(=-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(VP=3\left[xy+z\left(x+y+z\right)\right]\left(z-x-y-z\right)=3\left(xy+yz+zx+z^2\right)\left(-x-y\right)\)
\(=-3\left(y+z\right)\left(z+x\right)\left(x+y\right)\)
Do VT = VP nên ta có đpcm.