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a/ \(\dfrac{x^3}{x^2+1975}\cdot\dfrac{2x+1954}{x+1}+\dfrac{x^3}{x^2+1975}\cdot\dfrac{21-x}{x+1}=\dfrac{x^3\left(2x+1954\right)+x^3\left(21-x\right)}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{2x^4+1954x^3+21x^3-x^4}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{x^4+1975x^3}{\left(x^2+1975\right)\left(x+1\right)}\)
b/ \(\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{x^2+1945}\cdot\dfrac{x-2}{x-7}=\dfrac{\left(19x+8\right)\left(5x-9\right)+\left(19x+8\right)\left(x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{\left(19x+8\right)\left(5x-9+x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-209x+40x-88}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-169x-88}{x^2+1938x-13615}\)
c/ \(\dfrac{x+1}{x^2-2x-8}\cdot\dfrac{4-x}{x^2+x}=\dfrac{\left(x+1\right)\left(4-x\right)}{x\left[x^2-4x+2x-8\right]\left(x+1\right)}=-\dfrac{x-4}{x\left(x-4\right)+2\left(x-4\right)}=-\dfrac{x-4}{\left(x-4\right)\left(x+2\right)}=-\dfrac{1}{x+2}\)
\(\begin{array}{l}a)\dfrac{{{x^2} - 49}}{{{x^2} + 5}}.\left( {\dfrac{{{x^2} + 5}}{{x - 7}} - \dfrac{{{x^2} + 5}}{{x + 7}}} \right)\\ = \dfrac{{\left( {x - 7} \right)\left( {x + 7} \right)}}{{{x^2} + 5}}.\dfrac{{{x^2} + 5}}{{x - 7}} - \dfrac{{\left( {x - 7} \right)\left( {x + 7} \right)}}{{{x^2} + 5}}.\dfrac{{{x^2} + 5}}{{x + 7}}\\ = x + 7 - \left( {x - 7} \right) = 14\end{array}\)
\(\begin{array}{l}b)\dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 - x}}{{x + 1945}} + \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2{\rm{x}} - 25}}{{x + 1945}}\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\left( {\dfrac{{2000 - x}}{{x + 1945}} + \dfrac{{2{\rm{x}} - 25}}{{x + 1945}}} \right)\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{2000 - x + 2{\rm{x}} - 25}}{{x + 1945}}\\ = \dfrac{{19{\rm{x}} + 8}}{{x + 1975}}.\dfrac{{x + 1975}}{{x + 1945}} = \dfrac{{19{\rm{x}} + 8}}{{x + 1945}}\end{array}\)
a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)
Cách 1 \(\dfrac{x-1}{x}.\left(x^2+x+1+\dfrac{x^3}{x-1}\right)\\ =\dfrac{x-1}{x}.\left(\dfrac{\left(x-1\right)(x^2+x+1)+x^3}{x-1}\right)\\ =\dfrac{x-1}{x}.\dfrac{2x^3-1}{x-1}=\dfrac{2x^3-1}{x}\)
Cách 2 \(\dfrac{x-1}{x}.\left(x^2+x+1+\dfrac{x^3}{x-1}\right)\\ =\dfrac{x-1}{x}.\left(x^2+x+1\right)+\dfrac{x-1}{x}.\dfrac{x^3}{x-1}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x}+x^2\\ =\dfrac{x^3-1}{x}+x^2=\dfrac{2x^3-1}{x}\)
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(=\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x^2-x-2\right)}{x^2}\)
\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:
\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)
a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
=\(\dfrac{x^2+1-2x}{x-1}\)
=\(\dfrac{x^2-2x+1}{x-1}\)
=\(\dfrac{\left(x-1\right)^2}{x-1}\)
= x - 1
b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)
=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)
=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)
=\(\dfrac{x+3}{x}\)
#Fiona
