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30 tháng 7 2018
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
30 tháng 7 2018
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
PT
12 tháng 12 2017
a) \(\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{-1}{x-1}\)
\(=\dfrac{x^3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{-1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4-x+x^3+x+x-1-x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4+x^3}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3}{x-1}\)
b) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)-x^2\left(x-1\right)-1\left(x+1\right)+1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^3-x^3+x^2-x-1+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^2-2}{\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{4-2x+x^2}{2+x}-2-x\)
\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2\left(2+x\right)}{2+x}-\dfrac{x\left(2+x\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-4-2x-2x-x^2}{2+x}\)
\(=\dfrac{-6x}{2+x}\)
Còn lại thì dễ rồi, bạn tự làm nha ^^
ET
4 tháng 12 2018
a) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{15x.2y^2}{7y^3.x^2}=\dfrac{30}{7xy}\)
b) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)=\dfrac{-4y^2.3x^2}{11x^4.8y}=\dfrac{-3y}{22x^2}\)
c) \(\dfrac{x^3-8}{5x+20}.\dfrac{x^2+4x}{x^2+2x+4}\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{x^2-2x}{5}\)
Lời giải:
\(\frac{2}{x^2y}-\frac{4xy^2-1}{2x^3y^3}=\frac{4xy^2}{2x^3y^3}-\frac{4xy^2-1}{2x^3y^3}=\frac{4xy^2-(4xy^2-1)}{2x^3y^3}=\frac{1}{2x^3y^3}\)