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Lời giải:
Ta có:
\(x^3(y-z)+z^3(x-y)=y^3(x-z)=y^3[(y-z)+(x-y)]\)
\(\Leftrightarrow x^3(y-z)+z^3(x-y)-y^3(y-z)-y^3(x-y)=0\)
\(\Leftrightarrow (x^3-y^3)(y-z)-(y^3-z^3)(x-y)=0\)
\(\Leftrightarrow (x-y)(x^2+xy+y^2)(y-z)-(y-z)(y^2+yz+z^2)(x-y)=0\)
\(\Leftrightarrow (x-y)(y-z)(x^2+xy+y^2-y^2-yz-z^2)=0\)
\(\Leftrightarrow (x-y)(y-z)(x^2+xy-z^2-yz)=0\)
\(\Leftrightarrow (x-y)(y-z)(x-z)(x+y+z)=0\)
Vì $x,y,z$ đôi một khác nhau nên \((x-y)(y-z)(x-z)\neq 0\). Do đó $x+y+z=0$
Khi đó:
\(x^3+y^3+z^3=(x+y)^3-3xy(x+y)+z^3\)
\(=(-z)^3-3xy(-z)+z^3=-z^3+3xyz+z^3=3xyz\)
Ta có đpcm.
Ta có:
\(\frac{x}{1+x^2}+\frac{18y}{1+y^2}+\frac{4z}{1+z^2}=xyz\left(\frac{1}{yz\left(1+x^2\right)}+\frac{18}{xz\left(1+y^2\right)}+\frac{4}{xy\left(1+z^2\right)}\right)\)
\(=xyz\left(\frac{1}{yz+x\left(x+y+z\right)}+\frac{18}{xz+y\left(x+y+z\right)}+\frac{4}{xy+z\left(x+y+z\right)}\right)\)
\(=xyz\left(\frac{1}{\left(x+y\right).\left(x+z\right)}+\frac{18}{\left(y+x\right).\left(y+z\right)}+\frac{4}{\left(z+x\right).\left(z+y\right)}\right)\)
\(=xyz.\frac{\left(z+y\right)+18.\left(x+z\right)+4\left(x+y\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)
\(=\frac{xyz\left(22x+5y+19z\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)(đpcm)
Ta có : \(\left(x-y+z\right)^2=x^2-y^2+z^2\)
\(\Leftrightarrow x^2+y^2+z^2-2xy+2xz-2yz=x^2-y^2+z^2\)
\(\Leftrightarrow2y^2-2xy+2xz-2yz=0\)
\(\Leftrightarrow2y\left(y-z\right)-2x\left(y-z\right)=0\)
\(\Leftrightarrow2\left(y-x\right)\left(y-z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=z\end{matrix}\right.\)
Với x = y \(\Rightarrow\left(x-y+z\right)^n=z^n;x^n-y^n+z^n=z^n\)
\(\Rightarrow\left(x-y+z\right)^n=x^n-y^n+z^n\) ( 1 )
Với y = z \(\Rightarrow\left(x-y+z\right)^n=x^n;x^n-y^n+z^n=x^n\)
\(\Rightarrow\left(x-y+z\right)^n=x^n-y^n+z^n\) ( 2 )
Từ ( 1 ) ; ( 2 ) => ĐPCM