Làm biếng chép :'<
Link : Câu hỏi jj đó vào đây rồi biết :))
làm đc r mak, đăng lên thách đốmn tí thui
Ta có : \(\hept{\begin{cases}x=by+cz\\z=ax+by\end{cases}}< =>x-cz=z-ax< =>x\left(a+1\right)=z\left(c+1\right)\)
Tương tự ta cũng có : \(y\left(b+1\right)=x\left(a+1\right);z\left(c+1\right)=y\left(b+1\right)\)
Từ đó ta suy ra : \(x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)\)
Đặt \(k=x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)< =>3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)
Ta quay lại điều cần chứng minh : \(VT=\frac{x}{x\left(1+a\right)}+\frac{y}{y\left(1+b\right)}+\frac{z}{z\left(1+c\right)}=\frac{x}{k}+\frac{y}{k}+\frac{z}{k}\)
\(=\frac{3\left(x+y+z\right)}{3k}=\frac{3\left(x+y+z\right)}{x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)}=\frac{3\left(x+y+z\right)}{xa+yb+zc+x+y+z}\)
\(=\frac{3\left(x+y+z\right)}{\frac{1}{2}\left(xa+yb+cz+xa+yb+cz\right)+x+y+z}=\frac{3\left(x+y+z\right)}{\frac{3}{2}\left(x+y+z\right)}=2\left(đpcm\right)\)
Bài làm:
Ta có: \(\hept{\begin{cases}x=by+cz\\y=ax+cz\end{cases}}\Rightarrow x-y=by+cz-ax-cz\)
\(\Leftrightarrow x-y=by-ax\)
\(\Leftrightarrow x+ax=y+by\Leftrightarrow x\left(a+1\right)=y\left(b+1\right)\)
Tương tự CM được: \(x\left(a+1\right)=b\left(y+1\right)=c\left(z+1\right)\)
Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{x}{x\left(1+a\right)}+\frac{y}{y\left(b+1\right)}+\frac{z}{z\left(c+1\right)}\)
\(=\frac{x+y+z}{x\left(1+a\right)}=\frac{3\left(x+y+z\right)}{3x\left(1+a\right)}=\frac{3\left(x+y+z\right)}{x\left(1+a\right)+y\left(1+b\right)+z\left(1+c\right)}\)
\(=\frac{3\left(x+y+z\right)}{\left(ax+by+cz\right)+x+y+z}\)
\(=\frac{3\left(x+y+z\right)}{\frac{1}{2}\left(by+cz+ax+cz+ax+by\right)+x+y+z}\)\(=\frac{3\left(x+y+z\right)}{\frac{x+y+z}{2}+x+y+z}=\frac{3\left(x+y+z\right)}{\frac{3}{2}\left(x+y+z\right)}=\frac{3}{\frac{3}{2}}=2\)
