Ta có : \(y+z=ax+cz+ax+by=2ax+x\)

\(\Rightarrow\)\(y+z-x=2ax\)\(\Rightarrow\)\(a=\frac{y+z-x}{2x}\)\(\Rightarrow\)\(\frac{1}{a+1}=\frac{2x}{x+y+z}\)

Tương tự, ta cũng có \(\frac{1}{b+1}=\frac{2y}{x+y+z};\frac{1}{c+1}=\frac{2z}{x+y+z}\)

\(\Rightarrow\)\(S=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Chúc bạn học tốt ~ 

Theo đề ta có: xyz= 8.abc= xyz.abc= ax. by. cz= 8

                                                       hay ax.ax.ax= 8

=> (ax)³= 2³

=> ax= 2

Với ax= 2=> x= \(\frac{2}{a}\)

      by= 2=> y= \(\frac{2}{b}\)

      cz= 2=> z=\(\frac{2}{c}\)

Vậy x, y, z= \(\frac{2}{a},\frac{2}{b},\frac{2}{c}.\)

x.y.z= 8/ a.b.c =>abc.xyz=8 
a.x=b.y=c.z =>(a.x)^3=(b.y)^3=(c.z)^3 =ax.by.cz=8 
* (a.x)^3 =8 =>a.x=2 =>x=2/a 
* (b.y)^3 =8 =>b.y=2 =>y=2/b 
* (c.z)^3 =8 =>c.z=2=>z=2/c

\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2\left(abxy+bcyz+cazx\right)=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)\(\Leftrightarrow a^2y^2-2ay\cdot bx+b^2x^2+b^2z^2-2bz\cdot cy+c^2y^2+a^2z^2-2az\cdot cx+c^2x^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

mà \(\left(ay-bx\right)^2;\left(bz-cy\right)^2;\left(az-cx\right)^2\ge0\)nên \(\left(ay-bx\right)^2=\left(bz-cy\right)^2=\left(az-cx\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}ay=bx\\bz=cy\\az=cx\end{cases}\Leftrightarrow\frac{a}{x}}=\frac{b}{y}=\frac{c}{z}\left(x,y,z\ne0\right)\)(ĐPCM)

Bạn ko hiểu chỗ nào cứ hỏi lại mình nhé

x²-yz=a=>ax=x(x²-yz)=x³-xyz

tương tự và cộng lại ta có ax+by+cz=x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)=(x+y+z)(a+b+c) 

ta có đpcm