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a: \(=\dfrac{x^3-x^2-2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x}{\left(x-1\right)\left(x+1\right)}=x\)
b: \(=\dfrac{4x^2+4xy+y^2-8xy+4x^2-4xy+y^2}{\left(2x+y\right)\left(2x-y\right)}=\dfrac{12x^2-8xy+2y^2}{\left(2x-y\right)\left(2x+y\right)}\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
a) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{\left(2x+y\right)\left(2x+y\right)-8yx+\left(2x-y\right)\left(2x-y\right)}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{8x^2-8xy+2y^2}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(4x^2-4xy+y^2\right)}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(2x-y\right)^2}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(2x-y\right)}{x\left(2x+y\right)}\)
b) \(\dfrac{1}{x^2+3x+2}+\dfrac{2x}{x^2+4x+3}+\dfrac{1}{x^2+5x+6}\)
\(=\dfrac{1}{x^2+x+2x+2}+\dfrac{2x}{x^2+x+3x+3}+\dfrac{1}{x^2+2x+3x+6}\)
\(=\dfrac{1}{x\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{x\left(x+1\right)+3\left(x+1\right)}+\dfrac{1}{x\left(x+2\right)+2\left(x+2\right)}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x\left(x+2\right)+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x^2+4x+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2}{x+3}\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a: \(=\dfrac{x^2-1-3x^2+3+2x^2+7}{2x-y}=\dfrac{9}{2x-y}\)
b: \(=\dfrac{x+y+x-y+2x-3y}{1-xy}=\dfrac{4x-3y}{1-xy}\)
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{5}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)
\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
c) d) Tự làm đi mình làm biếng quass >.< ^^
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