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

Với a, b, c khác -1 thì x + y + z khác 0.
Từ đề bài ta có: y + z = ax + cz + ax + by
<=> 2ax = y + z - x
--> a = (y + z - x)/(2x) --> a + 1 = (x + y + z)/(2x)
--> 1/(1 + a) = 2x/(x + y + z)
tương tự: 1/(1 + b) = 2y/(x + y + z)
1/(1 + c) = 2z/(x + y + z)
--> 1/(1 + a) + 1/(1 + b) + 1/(1 + c) = (2x + 2y + 2z)/(x + y + z) = 2

vậy giá trị của biểu thức A= 2

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 ~ 

Ta có : \(\begin{cases}x=by+cz\\y=ax+cz\\z=ax+by\end{cases}\) . Cộng các đẳng thức trên theo vế :

\(x+y+z=2\left(ax+by+cz\right)\Rightarrow\frac{x+y+z}{ax+by+cz}=2\)

Lại có : \(y=ax+cz\Rightarrow a=\frac{y-cz}{x}\Rightarrow a+1=\frac{x+y-cz}{x}\Rightarrow\frac{1}{a+1}=\frac{x}{x+y-cz}=\frac{x}{ax+by+cz}\)

Tương tự : \(\frac{1}{b+1}=\frac{y}{ax+by+cz};\frac{1}{c+1}=\frac{z}{ax+by+cz}\)

\(\Rightarrow P=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x}{ax+by+cz}+\frac{y}{ax+by+cz}+\frac{z}{ax+by+cz}\)

\(=\frac{x+y+z}{ax+by+cz}=2\)

Ta có : \(\begin{cases}x=by+cz\\y=ax+cz\\z=ax+by\end{cases}\) . Cộng các đẳng thức trên theo vế : 

\(x+y+z=2\left(ax+by+cz\right)\)\(\Rightarrow\frac{x+y+z}{ax+by+cz}=2\)

Ta có : \(y=ax+cz\Rightarrow a=\frac{y-cz}{x}\Rightarrow a+1=\frac{x+y-cz}{x}\Rightarrow\frac{1}{a+1}=\frac{x}{x+y-cz}\)

\(\Rightarrow\frac{1}{a+1}=\frac{x}{ax+by+cz}\)

\(\Rightarrow P=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{x+y+z}{ax+by+cz}=2\)

Tương tự : \(\frac{1}{b+1}=\frac{y}{ax+by+cz}\) ; \(\frac{1}{c+1}=\frac{z}{ax+by+cz}\)

Với a, b, c khác -1 thì x + y + z khác 0.
Từ đề bài ta có: y + z = ax + cz + ax + by
<=> 2ax = y + z - x
--> a = (y + z - x)/(2x) --> a + 1 = (x + y + z)/(2x)
--> 1/(1 + a) = 2x/(x + y + z)
tương tự: 1/(1 + b) = 2y/(x + y + z)
1/(1 + c) = 2z/(x + y + z)
--> 1/(1 + a) + 1/(1 + b) + 1/(1 + c) = (2x + 2y + 2z)/(x + y + z) = 2

cộng 3 cái lại nhe bạn =))

Có nhiều cách làm bài này.

Có \(2a+2b+2c=by+cz+a.x+cz+a.x+by\)

\(2\left(a+b+c\right)=2\left(a.x+by+cz\right)\)

\(\Rightarrow a+b+c=a.x+by+cz\)

\(\Rightarrow\frac{1}{x+2}=\frac{a}{a+b+c}\)

\(\Rightarrow\frac{1}{z+2}=\frac{c}{a+b+c}\)

\(\Rightarrow\frac{1}{y+2}=\frac{b}{a+b+c}\)

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

Vậy ...

=1 bạn nhé