cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
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:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
cho x + y + z = 0 và x, y , z khác 0 hãy rút gọn
a) P = \(\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) Q = \(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
a)
Có \(x+y+z=0\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\) (1)
Phân tích :
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\)
\(=2\left(x^2+y^2+z^2\right)+\left[-2\left(xy+yz+xz\right)\right]\)(Áp dung (1)ta được :)
\(=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2\)
\(=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow P=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(\Rightarrow P=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\)
\(\Rightarrow P=\dfrac{1}{3}\)
Rút gọn: \(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\), biết rằng: x+y+z=0
\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(P=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}=\dfrac{1}{3}\)
Cho \(z^2+2\left(xy-xz-yz\right)=0,x+y\ne z,y\ne z\)
Chứng minh: \(\dfrac{x^2+\left(x-z\right)^2}{y^2+\left(y-z\right)^2}=\dfrac{x-z}{y-z}\)
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
Chứng minh đẳng thức:
a) \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}+\dfrac{z}{\left(y-z\right)\left(z-x\right)}+\dfrac{x}{\left(z-x\right)\left(x-y\right)=0}\)
b) \(\dfrac{x^2}{\left(x-y\right)\left(y-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)=1}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{1}{xyz}\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
CẦN GẤP!!
Cho x+y+z=0 xà x,y,z khác 0 rút gọn
a)P=\(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b)Q=\(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
\(x+y+z=0\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)
\(\Rightarrow P=\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left[-\left(y+z\right)\right]^2+\left[-\left(z+x\right)\right]^2+\left[-\left(x+y\right)\right]^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left(y+z\right)^2+\left(z+x\right)^2\left(x+y\right)^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{-\left[\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=-1\)
Mik mới biết làm câu a thôi còn câu b thì từ từ mik nghĩ đã nhé @-@
Chúc bn học giỏi nhoa!!!
Rút gọn:
A=\(\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)
Quy đồng tính bình thường.
\(A=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)\(=\dfrac{2x^2+2y^2+2z^2-2xy-2yz-2xz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\dfrac{2yz+2xz+2xy-2x^2-2y^2-2z^2}{ }\)
=0
