cho x=by+cz:y=ax+cz:z=ax+by và x+y+z#0,xyz=0 .cm đẳng thức 1/(1+a)+1/(1+b)+1/(1+c)=2
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27 tháng 7 2018
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
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NT
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22 tháng 10 2016
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 \(Q=2.\)
TN
12 tháng 12 2016
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
12 tháng 3 2021
Ta có:
\(2a+2b+2c=by+cz+ax+cz+ax+by\)
\(\Leftrightarrow a+b+c=ax+by+cz\)
\(\Rightarrow a+b+c=ax+2a;a+b+c=by+2b;a+b+c=cz+2c\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{a}{a+b+c};\frac{1}{y+2}=\frac{b}{a+b+c};\frac{1}{z+2}=\frac{c}{a+b+c}\)
\(\Rightarrow A=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
NC
12 tháng 3 2021
Ta có:\(\hept{\begin{cases}2a=by+cz\\2b=ax+cz\\2c=ax+by\end{cases}}\)
\(\Leftrightarrow2a+2b+2c=by+cz+ax+cz+ax+by\)
\(\Leftrightarrow2a+2b+2c=2ax+2by+2cz\)
\(\Leftrightarrow2a+2b+2c-2ax-2by-2cz=0\)
\(\Leftrightarrow\left(2a-2ax\right)+\left(2b-2by\right)+\left(2c-2cz\right)=0\)
\(\Leftrightarrow2a\left(1-x\right)+2b\left(1-y\right)+2c\left(1-z\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}1-x=0\\1-y=0\\1-z=0\end{cases}\Leftrightarrow x=y=z=1}\)
\(\Rightarrow A=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{1+2}+\frac{1}{1+2}+\frac{1}{1+2}=1\)