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 = \(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
1.Cho x+y+z=0 ,rút gọn:
\(A=\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
2.Tính \(A=\dfrac{x-y}{x+y}\)biết x2-2y2=xy (y khácx;x+y khác 0)
Cho x,y,z khác 0 và \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
CMR:\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Cho \(x+y+z=xyz\) và \(xy+yz+zx\ne-3\)
Chứng minh: \(\dfrac{x.\left(y^2+z^2\right)+y.\left(z^2+x^2\right)+z.\left(x^2+y^2\right)}{xy+yz+zx-3}=xyz\)
Giải PTNN:
a)\(x^2+2y^2+z^2-2xy-2y+2z+2=0\)
b)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{6xy}=\dfrac{1}{6}\)
c)\(x^3-y^3=xy+8\)
Chứng minh:
\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
a) \(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)
b) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
c) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
Bài 4: Chứng minh
\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
\(\dfrac{x}{\left(x+y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
Thực hiện phép tính:
1,\(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)
2,\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)
3,\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
4,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
5,\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)