Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cộng vế với vế:
\(\Rightarrow x+y+z=2ax+2by+2cz\)
\(\Rightarrow x+y+z-2x=2ax+2by+2cx-2\left(by+cz\right)=2ax\)
\(\Rightarrow2ax=y+z-x\)
\(\Rightarrow a=\dfrac{y+z-x}{2x}\Rightarrow1+a=\dfrac{x+y+z}{2x}\)
Tương tự ta có: \(1+b=\dfrac{x+y+z}{2y}\) ; \(1+c=\dfrac{x+y+z}{2z}\)
\(\Rightarrow\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=\dfrac{2x+2y+2z}{x+y+z}=2\)
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 \(Q=2.\)
ax + by = 5c (1); by + cz = 5a (2); cz + ax = 5b (3);
Lấy (1) - (2) + (3) về theo vế có : 2ax = - 5a + 5b + 5c => 2a(x + 5) = 5(a + b + c)
=> 1/(x + 5) = 2a/5(a + b + c) (4)
Tương tự :
1/(y + 5) = 2b/5(a + b + c) (5)
1/(z + 5) = 2c/5(a + b + c) (6)
Từ (4) + (5) + (6) :
M = 1/(x + 5) + 1/(y + 5) + 1/(z + 5) = 2/5
1 la sai ; 2 cung sai ; xin loi cho ming ting xiu ; aaaaa! 3 la ......................................sai; chan chan 4 la ..............................................................................................d...........................sai ; 1000000000000000000000000000000000000000000000000000000000000000000000000000 la ..................................................................................................sai
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
Ta có:
\(X-A\)\(=\)\(by+cz-cy-bz=\left(b-c\right)y+\left(c-b\right)z=\left(b-c\right)\left(y-z\right)\)
\(X-B\)\(=\)\(ax+by-bx-ay=\left(a-b\right)x+\left(b-a\right)y=\left(a-b\right)\left(x-y\right)\)
\(X-C\)\(=\)\(ax+cz-cx-az=\left(a-c\right)x+\left(c-a\right)z=\left(a-c\right)\left(x-z\right)\)
\(Y-A\)\(=\)\(cx+ay-ax-cy=\left(c-a\right)x+\left(a-c\right)y=\left(c-a\right)\left(x-y\right)\)
\(Y-B\)\(=\)\(cx+bz-bx-cz=\left(c-b\right)x+\left(b-c\right)z=\left(c-a\right)\left(x-z\right)\)
\(Y-C\)\(=\)\(zy+bz-by-az=\left(a-b\right)y+\left(b-a\right)z=\left(a-b\right)\left(y-z\right)\)
\(Z-A\)\(=\)\(bx+az-ax-bz=\left(b-a\right)x+\left(a-b\right)z=\left(b-a\right)\left(x-z\right)\)
\(Z-B\)\(=\)\(cy+az-ay-cz=\left(c-a\right)y+\left(a-c\right)z=\left(c-a\right)\left(y-z\right)\)
\(Z-C\)\(=\)\(bx+cy-cx-by=\left(b-c\right)x+\left(c-b\right)y=\left(b-c\right)\left(x-y\right)\)
Từ đó có:
\(\left(X-A\right)\left(X-B\right)\left(X-C\right)=\left(b-c\right)\left(a-b\right)\left(a-c\right)\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
\(\left(Y-A\right)\left(Y-B\right)\left(Y-C\right)=\left(c-a\right)\left(c-b\right)\left(a-b\right)\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
\(\left(Z-A\right)\left(Z-B\right)\left(Z-C\right)=\left(b-a\right)\left(c-a\right)\left(b-c\right)\left(x-z\right)\left(y-z\right)\left(x-z\right)\)
Ta thấy , vế phải của ba đẳng thức trên là tích của sáu thừa số . Các thừa số đều có mặt trong các tích nếu ta áp dụng quy tắc đổi dấu
có cần giải ra không