Câu hỏi 6.31
24
a) \(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\);
b) \(\frac{x}{2 x-y}+\frac{y}{2 x+y}+\frac{3 x y}{y^2-4 x^2}\)
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\(\begin{aligned} & \text { a) } \frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=\frac{z+x+y}{x y z} \text {; } \\ & \text { b) } \frac{x}{2 x-y}+\frac{y}{2 x+y}+\frac{3 x y}{y^2-4 x^2} \\ & =\frac{x}{2 x-y}+\frac{y}{2 x+y}-\frac{3 x y}{4 x^2-y^2} \\ & =\frac{x(2 x+y)+y(2 x-y)-3 x y}{(2 x+y)(2 x-y)} \\ & =\frac{2 x^2+x y+2 x y-y^2-3 x y}{(2 x+y)(2 x-y)} \\ & =\frac{2 x^2-y^2}{(2 x+y)(2 x-y)} . \end{aligned}\begin{aligned} & \text { a) } \frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=\frac{z+x+y}{x y z} \text {; } \\ & \text { b) } \frac{x}{2 x-y}+\frac{y}{2 x+y}+\frac{3 x y}{y^2-4 x^2} \\ & =\frac{x}{2 x-y}+\frac{y}{2 x+y}-\frac{3 x y}{4 x^2-y^2} \\ & =\frac{x(2 x+y)+y(2 x-y)-3 x y}{(2 x+y)(2 x-y)} \\ & =\frac{2 x^2+x y+2 x y-y^2-3 x y}{(2 x+y)(2 x-y)} \\ & =\frac{2 x^2-y^2}{(2 x+y)(2 x-y)} . \end{aligned}\)