\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-xz=b\\z^2-xy=c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\) \(\Rightarrow ax+by+cz=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)⋮\left(x+y+z\right)\)

Ta có ax + by = c ; by + cz = a

<=> cz - ax = a - c (1)

mà cz + ax = b (2) 

Từ (1) và (2) => \(cz=\frac{a-c+b}{2}\Rightarrow z=\frac{a-c+b}{2c}\Rightarrow z+1=\frac{a+b+c}{2c}\)

=> \(\frac{1}{z+1}=\frac{2c}{a+b+c}\)

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

=> P = \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=2\)

Vì \(x=by+cz\)

\(\Rightarrow by=x-cz\)

Mà \(z=ax+by\)

\(\Rightarrow by=z-ax\)

\(\Rightarrow x-cz=z-ax\left(=by\right)\)

\(\Rightarrow x+ax=z+cz\)

\(\Rightarrow x\left(a+1\right)=z\left(c+1\right)\)

Cũng có :

\(z=ax+by\)

\(\Rightarrow ax=z-by\)

\(y=ax+cz\)

\(\Rightarrow ax=y-cz\)

\(\Rightarrow z-by=y-cz\left(=ax\right)\)

\(\Rightarrow z+cz=y+by\)

\(\Rightarrow z\left(c+1\right)=y\left(b+1\right)\)

\(\Rightarrow x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)\)

Đặt \(x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)=k\)

\(\Rightarrow3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)

Có :

\(Q=\frac{1}{a+1}+\frac{1}{1+b}+\frac{1}{c+1}\)

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

\(=\frac{x}{k}+\frac{y}{k}+\frac{z}{k}\)

\(=\frac{x+y+z}{k}\)

\(=\frac{3\left(x+y+z\right)}{3k}\)

Mà \(3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)

\(\Rightarrow Q=\frac{3\left(x+y+z\right)}{x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)}\)

\(=\frac{3\left(x+y+z\right)}{xa+x+by+y+zc+z}\)

\(=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\left(xa+by+zc\right)}\)

\(=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\frac{1}{2}\left[\left(xa+by\right)+\left(xa+zc\right)+\left(by+zc\right)\right]}\)

Có \(x+y+z=\left(ax+by\right)+\left(by+cz\right)+\left(ax+cz\right)\)

\(\Rightarrow Q=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\frac{1}{2}\left(x+y+z\right)}\)

\(=\frac{3\left(x+y+z\right)}{\frac{3}{2}\left(x+y+z\right)}\)

\(=\frac{3}{\frac{3}{2}}\)

\(=2\)

Vậy \(Q=2.\)

Tim x toa man: |x-22|+|x-3|+|x-2017|=2014

Ta có

x²-yz=a

y²-zx=b

z²-xy=c

=>x³-xyz=ax

    y³-xyz=by

    z³-xyz=cz

=> x³+y³+z³-3xyz=ax+by+cz

Lại có

x³+y³+z³-3xyz

=(x+y)³-3x²y-3xy²+z³-3xyz

=[(x+y)³+z³]-3xy(x+y+z)

Áp dụng hằng đẳng thức x³+y³=(x+y)(x²-xy+y²) ta được:

=(x+y+z)[(x+y)²-z(x+y)+z²]-3xy(x+y+z)

=(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)

=(x+y+z)(x²+y²+z²-xy-yz-zx)

( Hình như phải Chứng minh ax+by+cz chia hết cho x+y+z chứ nhỉ, nếu ko phải thì cho mik srr nhé, nếu đúng như mình nói thì bạn làm như trên nha)

ak mình nhầm tẹo srr nha, đến chỗ

(x+y+z)(x²+y²+z²-xy-yz-zx)

Vì x²-yz=a, y²-zx=b, z²- xy=c

=>x²+y²+z²-xy-yz-zx=a+b+c

=>ax+by+cz=(x+y+z)(a+b+c)

=> DPCM

srr nhiều

\(2x-2y=by+cz-cz-ax=by-ax\)

\(\Rightarrow2x-2y=by-ax\)

\(\Rightarrow2x+ax=2y+by\)

\(\Rightarrow x\left(a+2\right)=y\left(b+2\right)\)

\(\Rightarrow a+2=\dfrac{y\left(b+2\right)}{x}\)

\(2z-2y=ax+by-cz-ax=by-cz\)

\(\Rightarrow2z+cz=2y+by\)

\(\Rightarrow z\left(c+2\right)=y\left(b+2\right)\)

\(\Rightarrow c+2=\dfrac{y\left(b+2\right)}{z}\)

\(A=\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}=\dfrac{2}{\dfrac{y\left(b+2\right)}{x}}+\dfrac{2}{b+2}+\dfrac{2}{\dfrac{y\left(b+2\right)}{z}}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2}{b+2}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2y}{y\left(b+2\right)}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x+2y+2z}{y\left(b+2\right)}=\dfrac{by+cz+cz+ax+ax+by}{by+2y}=\dfrac{2\left(ax+by+cz\right)}{by+cz+ax}=2\)