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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: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
dài lắm
\(\left(x+y\right):\left(8-z\right):\left(y+z\right):\left(10+z\right)=2:5:3:4\\ < =>\dfrac{x+y}{2}=\dfrac{8-z}{5}=\dfrac{y+z}{3}=\dfrac{10-z}{4}\left(1\right)\)
\(\left(1\right)=>\dfrac{8-z}{5}=\dfrac{10+z}{4}\\ < =>4\left(8-z\right)=5\left(10+z\right)\\ < =>32-4z=50+5z\\ < =>-9z=18\\ < =>z=-2\left(2\right)\)
\(\left(1\right)=>\dfrac{y+z}{3}=\dfrac{8-z}{5}\left(3\right)\)
thay (2) vào (3)
\(=>\dfrac{y-2}{3}=\dfrac{8+2}{5}\\ < =>\dfrac{y-2}{3}=2\\ < =>y=8\left(4\right)\)
\(\left(1\right)=>\dfrac{x+y}{2}=\dfrac{8-z}{5}\left(5\right)\)
thay 4 và 2 vào 5
\(=>\dfrac{x+8}{2}=\dfrac{8+2}{5}\\ < =>\dfrac{x+8}{2}=2\\ < =>x=-4\left(6\right)\)
\(=>\dfrac{xyz}{x+y+z}\\ =\dfrac{\left(-2\right).8.\left(-4\right)}{\left(-4\right)+8+\left(-2\right)}\\ =\dfrac{64}{2}\\ =32\)
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