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a) P = (\(\dfrac{1}{2x-y}\) + \(\dfrac{3y}{y^2-4x^2}\) - \(\dfrac{2}{2x+y}\) )
= (\(\dfrac{1}{2x-y}\) - \(\dfrac{3y}{4x^2-y^2}\) - \(\dfrac{2}{2x+y}\) )
= ( \(\dfrac{2x+y}{\left(2x-y\right)\left(2x+y\right)}\) - \(\dfrac{3y}{\left(2x+y\right)\left(2x-y\right)}\) - \(\dfrac{2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)}\) )
= \(\dfrac{2x+y-3y-4x+2y}{\left(2x-y\right)\left(2x+y\right)}\) = \(\dfrac{-2x}{\left(2x+y\right)\left(2x-y\right)}\)
a: \(\left(\dfrac{1}{\left(2x-y\right)^2}+\dfrac{2}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{1}{\left(2x+y\right)^2}\right)\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{4x^2+4xy+y^2+2\left(4x^2-y^2\right)+4x^2-4xy+y^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{8x^2+2y^2+8x^2-2y^2}{\left(2x-y\right)^2}\cdot\dfrac{1}{16x}\)
\(=\dfrac{16x^2}{16x}\cdot\dfrac{1}{\left(2x-y\right)^2}=\dfrac{x}{\left(2x-y\right)^2}\)
b: \(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)
\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\dfrac{2x}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{-x}=\dfrac{-2\left(x-2\right)}{x+2}\)
câu này post hồi học lớp 8 = )) giờ tốt nghiệp c3 thì có người trả lời :'))
khbiet nên cười hay khóc đây
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
Đặt \(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+x+xy+y}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(A=\left[\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-1}{\left(x+y\right).\left(x-2y\right)}\right):\dfrac{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}{\left(x+y\right).\left(x+1\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)
\(A=\left(\dfrac{\left(x-y\right).\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right).\left(2y-x\right)}.\dfrac{\left(x+y\right).\left(x+1\right)}{\left(2x^2+y+2\right).\left(2x^2+y-2\right)}\right):\dfrac{2x^2+y+2}{x+1}\)
\(A=\left(\dfrac{2x^2+y-2}{2y-x}.\dfrac{x+1}{2x^2+y-2}\right).\dfrac{1}{x+1}\)
\(A=\dfrac{1}{2y-x}\)
Thay \(x=-1,76\) và \(y=\dfrac{3}{25}\) vào biểu thức ta được:
\(A=\dfrac{1}{2.\dfrac{3}{25}-\left(-1,76\right)}\)
\(A=\dfrac{1}{2}\)
a) điều kiện : \(x\ne\pm\dfrac{y}{2}\)
ta có : \(P=\left(\dfrac{1}{2x-y}+\dfrac{3y}{y^2-4x^2}-\dfrac{2}{2x+y}\right):\left(\dfrac{4x^2+y^2+1}{4x^2-y^2}\right)\)
\(\Leftrightarrow P=\left(\dfrac{1}{2x-y}-\dfrac{3y}{4x^2-y^2}-\dfrac{2}{2x+y}\right):\left(\dfrac{4x^2+y^2+1}{4x^2-y^2}\right)\)
\(\Leftrightarrow P=\left(\dfrac{1}{2x-y}-\dfrac{3y}{\left(2x-y\right)\left(2x+y\right)}-\dfrac{2}{2x+y}\right):\left(\dfrac{4x^2+y^2+1}{4x^2-y^2}\right)\)
\(\Leftrightarrow P=\left(\dfrac{2x+y-3y-2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)}\right).\left(\dfrac{4x^2-y^2}{4x^2+y^2+1}\right)\) \(\Leftrightarrow P=\left(\dfrac{-2x}{\left(2x-y\right)\left(2x+y\right)}\right).\left(\dfrac{\left(2x-y\right)\left(2x+y\right)}{4x^2+y^2+1}\right)\) \(\Leftrightarrow P=\dfrac{-2x}{4x^2+y^2+1}\) b) đề thế này thì làm không đc câu b nha bn , bn xem lại đề thử có sai không