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Lời giải:
Nếu $x+y+z=0$ thì:
$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$
$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$
$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$
(thỏa mãn đkđb)
Khi đó:
$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$
$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$
Nếu $x+y+z\neq 0$
Áp dụng TCDTSBN:
$\frac{x+y-z}{z}=\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z+y+z-x+z+x-y}{z+x+y}=\frac{x+y+z}{x+y+z}=1$
$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:
$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$
\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)
<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)
<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)
đặt x+y+z = t
=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)
=> x=y=z
ta lại có
\(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)
vì x=y=z => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
a) Với \(x+y+z=0\) ta tìm được \(\left(x;y;z\right)\rightarrow\left(0;0;0\right)\)
Với \(x+y+z\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)
Hay: \(x+y+z=\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}y+z=\dfrac{1}{2}-x\\x+z=\dfrac{1}{2}-y\\x+y=\dfrac{1}{2}-z\end{matrix}\right.\)
Thay vào đề bài ta được:
\(\dfrac{x}{\dfrac{1}{2}-x+1}=\dfrac{y}{\dfrac{1}{2}-y+1}=\dfrac{z}{\dfrac{1}{2}-z-2}=\dfrac{1}{2}\) Dễ dàng tìm được x;y;z
b) Theo đề bài ta có sẵn x+y+z khác 0
Áp dụng dãy tỉ số rồi làm tương tự câu a
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
Lời giải:
Áp dụng TCDTSBN:
$\frac{1}{x+y+z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2(x+y+z)}{x+y+z}=2$
\(\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ y+z+1=2x\\ x+z+2=2y\\ x+y-3=2z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y+z=\frac{1}{2}\\ x+y+z+1=3x\\ x+y+z+2=3y\\ x+y+z-3=3z\end{matrix}\right.\)
\(\left\{\begin{matrix} \frac{1}{2}+1=3x\\ \frac{1}{2}+2=3y\\ \frac{1}{2}-3=3z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{5}{6}\\ z=\frac{-5}{6}\end{matrix}\right.\)
Lời giải:
\(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(A+3=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{x+y}+1\right)\)
\(A+3=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\)
\(A+3=2017\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(A+3=2017.\frac{1}{672}=\frac{2017}{672}\)
\(\Rightarrow A=\frac{2017}{672}-3=\frac{1}{672}\)
Có
z= 2016 - (x+ y)
x= 2016- (y+ z)
y= 2016- (x+ z)
Thế vào ta được
[2016 - (x+ y)]/ (x+ y) + [2016 - (z+ y)]/ (z+ y) + [2016 - (x+ z)]/ (x+ z)
=2016/ (x+ y) - 1 + ( 2016)/ (z+ y)- 1 + ( 2016)/ (x+ z)-1
= 2016/ 8 - 3= 249