Cho x = by + cz ; y = ax + cz; z = ax + by.
\(CMR:A=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=2\)
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NU
0
NT
2
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 \(Q=2.\)
G
0
H
3
27 tháng 7 2018
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S
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
HK
1
Ta có
\(x-y=\left(by+cz\right)-\left(ax+cz\right)=by-ax\)
\(\Leftrightarrow x\cdot\left(a+1\right)=y\cdot\left(b+1\right)\)
\(y-z=\left(ax+cz\right)-\left(ax+by\right)=cz-by\)
\(\Leftrightarrow z\cdot\left(c+1\right)=y\cdot\left(b+1\right)\)
\(x-z=\left(by+cz\right)-\left(ax+by\right)=cz-ax\)
\(\Leftrightarrow x\cdot\left(a+1\right)=z\cdot\left(c+1\right)\)
\(\Rightarrow x\cdot\left(a+1\right)=z\cdot\left(c+1\right)=y\left(b+1\right)\)
Đặt \(x\cdot\left(a+1\right)=z\cdot\left(c+1\right)=y\left(b+1\right)=k\)
\(\Rightarrow\left\{{}\begin{matrix}a+1=\dfrac{k}{x}\\b+1=\dfrac{k}{y}\\c+1=\dfrac{k}{z}\end{matrix}\right.\)
Thay vào A, ta có :
\(A=\dfrac{1}{\dfrac{k}{x}}+\dfrac{1}{\dfrac{k}{y}}+\dfrac{1}{\dfrac{k}{z}}\)
\(=\dfrac{x}{k}+\dfrac{y}{k}+\dfrac{z}{k}\)
=\(\dfrac{x+y+z}{k}\)
Vì z = ax + by; x = cz + by; y = ax + cz nen :
\(k=z\cdot\left(c+1\right)=cz+z=cz+ax+by\)
\(\Rightarrow A=\dfrac{2\cdot\left(ax+by+czz\right)}{ax+by+cz}=2\)
⇒ĐPCM