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\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}=\frac{\left(x-z\right)-\left(x-y\right)}{\left(x-y\right)\left(x-z\right)}=\frac{1}{x-y}-\frac{1}{x-z}\)
\(\frac{z-x}{\left(y-z\right)\left(y-x\right)}=\frac{\left(y-x\right)-\left(y-z\right)}{\left(y-z\right)\left(y-x\right)}=\frac{1}{y-z}-\frac{1}{y-x}\)
\(\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{\left(z-y\right)-\left(z-x\right)}{\left(z-x\right)\left(z-y\right)}=\frac{1}{z-x}-\frac{1}{z-y}\)
Suy ra: \(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{1}{x-y}-\frac{1}{x-z}+\frac{1}{y-z}-\frac{1}{y-x}+\frac{1}{z-x}-\frac{1}{z-y}\)
\(=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
rồi bí mẹ chỗ này
\(A=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z-xy^2+y^2z+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left[xy-zx-zy+z^2\right]\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Vậy A = 1
Đặt \(\frac{x}{2013}=\frac{y}{2014}=\frac{z}{2015}=k\)
\(\Rightarrow\hept{\begin{cases}x=2013k\\y=2014k\\z=2015k\end{cases}}\)
Ta có :
4(x - y)(y - z) = 4(2013k - 2014k)(2014k - 2015k)
=4.(-k).(-k) = 4k2 (1)
(z - x)2 = (2015k - 2013k)2 = (2k)2 = 4k2 (2)
Từ 1 và 2
=> 4(x - y)(y - z) = (z - x)2
P=\(\frac{x}{\left(x-y\right)\left(x-z\right)}+\frac{y}{\left(y-x\right)\left(y-z\right)}+\frac{z}{\left(z-y\right)\left(z-x\right)}\) =\(\frac{x}{\left(x-y\right)\left(x-z\right)}-\frac{y}{\left(x-y\right)\left(y-z\right)}+\frac{z}{\left(y-z\right)\left(x-z\right)}\) =\(\frac{x\left(y-z\right)-y\left(x-z\right)+z\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\) =\(\frac{xy-xz-xy+yz+xz-yz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\) =0
+ Nếu x + y + z = 0 => x + y = -z; y + z = -x; x + z = -y
A = (1 + y/x)(1 + z/y)(1 + x/z)
A = (x+y)/x . (y+z)/y . (x+z)/z
A = -z/x . (-x)/y . (-y)/z = -1
+ Nếu x + y + z khác 0
x-y-z/x = -x+y-z/y = -x-y+z/z
<=> 1 - (y+z)/x = 1 - (x+z)/y = 1 - (x+y)/z
<=> y+z/x = x+z/y = x+y/z
Áp dụng t/c của dãy tỉ số = nhau ta có:
y+z/x = x+z/y = x+y/z = 2(x+y+z)/x+y+z = 2
A = (x+y)/x . (y+z)/y . (x+z)/z = 8
Bạn quy đồng rồi phân tích tử thành nhân tử rồi ra à.