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Lời giải:
Ta có:
\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x-y=by-ax\\ z=ax+by\end{matrix}\right.\)
\(\Rightarrow x-y+z=2by\Rightarrow b=\frac{x+z-y}{2y}\)
Hoàn toàn tương tự ta nhận được:
\(a=\frac{y+z-x}{2x};c=\frac{x+y-z}{2z}\)
Suy ra:
\(\left\{\begin{matrix} a+1=\frac{x+y+z}{2x}\\ b+1=\frac{x+y+z}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\) (ĐPCM)
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 \(Q=2.\)
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
Có nhiều cách làm bài này.
Có \(2a+2b+2c=by+cz+a.x+cz+a.x+by\)
\(2\left(a+b+c\right)=2\left(a.x+by+cz\right)\)
\(\Rightarrow a+b+c=a.x+by+cz\)
- \(a+b+c=a.x+\left(by+cz\right)=a.x+2.a=a\left(x+2\right)\)
\(\Rightarrow\frac{1}{x+2}=\frac{a}{a+b+c}\)
- \(a+b+c=\left(a.x+by\right)+cz=2c+cz=c\left(z+2\right)\)
\(\Rightarrow\frac{1}{z+2}=\frac{c}{a+b+c}\)
- \(a+b+c=by+\left(a.x+cz\right)=by+2b=b\left(y+2\right)\)
\(\Rightarrow\frac{1}{y+2}=\frac{b}{a+b+c}\)
\(\Rightarrow M=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{a+b+c}{a+b+c}=1\)
Vậy ...
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x+y+z=0 sao tính được. sửa đề: x+y+z khác 0
Ta có: \(x+y=by+cz+ax+cz=2cz+z\Leftrightarrow2cz=x+y-z\Leftrightarrow c=\frac{x+y-z}{2z}\Leftrightarrow c+1=\frac{x+y+z}{2z}\Leftrightarrow\frac{1}{c+1}=\frac{2z}{x+y+z}\left(1\right)\)
Tương tự, ta có: \(\frac{1}{a+1}=\frac{2x}{x+y+z}\left(2\right);\frac{1}{b+1}=\frac{2y}{x+y+z}\left(3\right)\)
Cộng (1),(2),(3) vế với vế ta được:
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2\left(x+y+z\right)}{x+y+z}=2\) hay Q = 2
Vậy Q=2
x=by+cz,y=ax+cz,z=ax+by
=>x+y+z=2(ax+by+cz) (1)
Thay z=ax+by vào (1) ta có :
x+y+z=2(z+cz)=2z(c+1)
\(=>\frac{1}{c+1}=\frac{2z}{x+y+z}\)
Tương tự ta có : \(\frac{1}{a+1}=\frac{2x}{x+y+z},\frac{1}{b+1}=\frac{2y}{x+y+z}\)
=>Q=\(\frac{2\left(x+y+z\right)}{x+y+z}=2\)