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\(a)\dfrac{{4{\rm{x}} + 3y}}{{{x^2} - {y^2}}} - \dfrac{{3{\rm{x}} + 4y}}{{{x^2} - {y^2}}} = \dfrac{{\left( {{\rm{4x}} + 3y} \right) - \left( {3{\rm{x}} + 4y} \right)}}{{{x^2} - {y^2}}} = \dfrac{{4{\rm{x}} + 3y - 3{\rm{x}} - 4y}}{{{x^2} - {y^2}}} = \dfrac{{x - y}}{{{x^2} - {y^2}}} = \dfrac{{x - y}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{1}{{x + y}}\)
\(\begin{array}{l}b)\dfrac{{2{\rm{x}}y - 3{y^2}}}{{{x^2} - 3{\rm{x}}y}} - \dfrac{x}{{3{\rm{x}} - 9y}}\\ = \dfrac{{2{\rm{x}}y - 3{y^2}}}{{x\left( {x - 3y} \right)}} - \dfrac{{{x^2}}}{{3\left( {x - 3y} \right)}}\\ = \dfrac{{3\left( {2{\rm{x}}y - 3{y^2}} \right)}}{{3{\rm{x}}\left( {x - 3y} \right)}} - \dfrac{{{x^2}}}{{3{\rm{x}}\left( {x - 3y} \right)}}\\ = \dfrac{{6{\rm{x}}y - 9{y^2} - {x^2}}}{{3{\rm{x}}\left( {x - 3y} \right)}} = \dfrac{{ - \left( {{x^2} - 6{\rm{x}}y + 9{y^2}} \right)}}{{3{\rm{x}}\left( {x - 3y} \right)}} = \dfrac{{ - {{\left( {x - 3y} \right)}^2}}}{{3{\rm{x}}\left( {x - 3y} \right)}} = \dfrac{{ - \left( {x - 3y} \right)}}{{3{\rm{x}}}}\end{array}\)
\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right).2\left( {x - 2} \right)}}{{4.\left( {x - 2} \right).\left( {x + 2} \right)}} = \dfrac{3}{2}\)
\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}} = \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)\left( {x + 5} \right)}}{{2\left( {x + 5} \right).\left( { - 1} \right)\left( {x - 6} \right)}} = \dfrac{{x + 6}}{{ - 2}} = \dfrac{{-x- 6}}{{ 2}}\)
\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}} = \dfrac{{\left( {1 - y} \right)\left( {1 + y + {y^2}} \right).5\left( {y + 1} \right)}}{{\left( {y + 1} \right).\left( {{y^2} + y + 1} \right)}} = 5\left( {1 - y} \right)\)
\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right) = \dfrac{{\left( {x + 2y} \right).\left( {2{\rm{x}} - y} \right)}}{{{{\left( {2{\rm{x}} - y} \right)}^2}}} = \dfrac{{x + 2y}}{{2{\rm{x}} - y}}\)
a)
\(\dfrac{{5{\rm{x}} - 4}}{9} + \dfrac{{4{\rm{x}} + 4}}{9} \\= \dfrac{{5{\rm{x}} - 4 + 4{\rm{x}} + 4}}{9} \\= \dfrac{{9{\rm{x}}}}{9} \\= x\)
b)
\(\dfrac{{{x^2}y - 6}}{{2{{\rm{x}}^2}y}} + \dfrac{{6 - x{y^2}}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{{x^2}y - 6 + 6 - x{y^2}}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{{x^2}y - x{y^2}}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{xy\left( {x - y} \right)}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{x - y}}{{2{\rm{x}}}}\)
c)
\(\dfrac{{x + 1}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x - 18}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x + 2}}{{{x^2} - 5{\rm{x}}}} \\= \dfrac{{x + 1 + x - 18 + x + 2}}{{{x^2} - 5{\rm{x}}}} \\= \dfrac{{3{\rm{x}} - 15}}{{x\left( {x - 5} \right)}} \\= \dfrac{{3\left( {x - 5} \right)}}{{x\left( {x - 5} \right)}} \\= \dfrac{3}{x}\)
d)
\(\dfrac{{7y}}{3} - \dfrac{{7y - 5}}{3} \\= \dfrac{{7y - 7y + 5}}{3} \\= \dfrac{5}{3}\)
e)
\(\dfrac{{4{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} - \dfrac{{7{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} \\= \dfrac{{4{\rm{x}} - 1 - 7{\rm{x}} + 1}}{{3{\rm{x}}{y^2}}} \\= \dfrac{{-3{\rm{x}}}}{{3{\rm{x}}{y^2}}} \\= \dfrac{-1}{{{y^2}}}\)
g)
\(\dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} - \dfrac{{x - y}}{{2y - x}} \\= \dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} + \left( { - \dfrac{{x - y}}{{2y - x}}} \right) \\= \dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} + \dfrac{{x - y}}{{x - 2y}} \\= \dfrac{{3y - 2{\rm{x}} + x - y}}{{x - 2y}} \\= \dfrac{{2y - x}}{{ - \left( {2y - x} \right)}} \\= - 1\)
\(\begin{array}{l}a)\dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{{x^2} - 4}}{{x + 1}}.\dfrac{{x - 2}}{{y + 6}}\\ = \dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{x - 2}}{{y + 6}}.\dfrac{{{x^2} - 4}}{{x + 1}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 4{\rm{x}} + 4} \right).\left( {y + 6} \right).\left( {x + 1} \right)}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {x - 2} \right)\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}.\left( {y + 6} \right).\left( {x + 1} \right)}} = \dfrac{{x + 2}}{{x + 1}}\end{array}\)
\(\begin{array}{l}b)\left(\frac{2x+1}{{x - 3}} + \frac{2x+1}{x+3}\right ) .\dfrac{{x^2 - 9}}{{2{\rm{x}} + 1}} \\ = (2x+1) \left ( \frac {1}{x-3} + \frac {1}{x+3} \right ) . \frac {(x-3)(x+3)}{2x + 1} \\ = (2x+1) \frac {x+3 + x - 3}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x + 1} \\ = \frac {2x(2x+1)}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x +1} \\= 2x \end{array}\)
\(a)\dfrac{{20{\rm{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{\rm{x}}^2}}}{{6y}}} \right) = \dfrac{{20{\rm{x}}}}{{3{y^2}}}.\left( { - \dfrac{{6y}}{{15{{\rm{x}}^2}}}} \right) = \dfrac{{20{\rm{x}}.\left( { - 6y} \right)}}{{3{y^2}.15{{\rm{x}}^2}}} = \dfrac{{ - 8}}{{3{\rm{x}}y}}\)
\(b)\dfrac{{9{{\rm{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{\rm{x}} + y}}{{2{\rm{x}} + 2y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right)}}{{x + y}}.\dfrac{{2{\rm{x}} + 2y}}{{3{\rm{x}} + y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right).2.\left( {x + y} \right)}}{{(x + y).\left( {3{\rm{x}} + y} \right)}} = 2\left( {3{\rm{x}} - y} \right)\)
\(\begin{array}{l}c)\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}} = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{y - x}}.\dfrac{{{x^2} - 2{\rm{x}}y + {y^2}}}{{{x^2} - xy + {y^2}}}\\ = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right).{{\left( {x - y} \right)}^2}}}{{ - (x - y)\left( {{x^2} - xy + {y^2}} \right)}} = \left( {x + y} \right)\left( {y - x} \right) = {{y^2} - {x^2}} \end{array}\)
\(d)\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right) = \dfrac{{\left( {3 - x} \right)\left( {3 + x} \right)}}{x}.\dfrac{1}{{x - 3}} = \dfrac{{ - \left( {x - 3} \right)\left( {3 + x} \right)}}{{x.\left( {x - 3} \right)}} = \dfrac{{ - \left( {3 + x} \right)}}{x}.\)
\(\begin{array}{l}a)\dfrac{{4{\rm{x}} + 2}}{{4{\rm{x - 4}}}} + \dfrac{{3 - 6{\rm{x}}}}{{6{\rm{x}} - 6}} = \dfrac{{2\left( {2x + 1} \right)}}{{4\left( {x - 1} \right)}} + \dfrac{{3\left( {1 - 2x} \right)}}{{6\left( {x - 1} \right)}}\\ = \dfrac{{2x + 1}}{{2\left( {x - 1} \right)}} + \dfrac{{1 - 2x}}{{2\left( {x - 1} \right)}} = \dfrac{{2x + 1 + 1 - 2x}}{{2\left( {x - 1} \right)}} = \dfrac{2}{{2\left( {x - 1} \right)}} = \dfrac{1}{{x - 1}}\end{array}\)
\(\begin{array}{l}b)\dfrac{y}{{2{{\rm{x}}^2} - xy}} + \dfrac{{4{\rm{x}}}}{{{y^2} - 2{\rm{x}}y}} = \dfrac{y}{{x\left( {2{\rm{x}} - y} \right)}} + \dfrac{{4{\rm{x}}}}{{y\left( {y - 2{\rm{x}}} \right)}}\\ = \dfrac{y}{{x\left( {2{\rm{x}} - y} \right)}} - \dfrac{{4{\rm{x}}}}{{y\left( {2{\rm{x}} - y} \right)}} = \dfrac{{{y^2}}}{{xy\left( {2{\rm{x}} - y} \right)}} - \dfrac{{4{{\rm{x}}^2}}}{{xy\left( {2{\rm{x}} - y} \right)}}\\ = \dfrac{{{y^2} - 4{{\rm{x}}^2}}}{{xy\left( {2{\rm{x}} - y} \right)}} = \dfrac{{\left( {y - 2{\rm{x}}} \right)\left( {y + 2{\rm{x}}} \right)}}{{ - xy\left( {y - 2{\rm{x}}} \right)}} = \dfrac{{ - \left( {y + 2{\rm{x}}} \right)}}{{xy}}\end{array}\)
\(\begin{array}{l}c)\dfrac{x}{{x - y}} + \dfrac{y}{{x + y}} + \dfrac{{2{y^2}}}{{{x^2} - {y^2}}}\\ = \dfrac{x}{{x - y}} + \dfrac{y}{{x + y}} + \dfrac{{2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\ = \dfrac{{x\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}} + \dfrac{{y\left( {x - y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}} + \dfrac{{2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\ = \dfrac{{{x^2} + xy + {\rm{yx}} - {y^2} + 2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{{x^2} + 2{\rm{x}}y + {y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{x + y}}{{x - y}}\end{array}\)
\(\begin{array}{l}d)\dfrac{{x{}^2 + 2}}{{{x^3} - 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{1}{{1 - x}}\\ = \dfrac{{x{}^2 + 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{x}{{{x^2} + x + 1}} - \dfrac{1}{{x - 1}}\\ = \dfrac{{x{}^2 + 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \dfrac{{{x^2} + 2 + {x^2} - x - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{{x^2} - 2{\rm{x}} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{x - 1}}{{{x^2} + x + 1}}\end{array}\)
a) MTC chọn là: \(2{{\rm{x}}^2}{y^4}\)
Nhân tử phụ của \(\dfrac{5}{{2{{\rm{x}}^2}{y^3}}}\) và \(\dfrac{3}{{x{y^4}}}\) lầm lượt là: y; 2x
Vậy: \(\begin{array}{l}\dfrac{5}{{2{{\rm{x}}^2}{y^3}}} = \dfrac{{5.y}}{{2{{\rm{x}}^2}{y^3}.y}} = \dfrac{{5y}}{{2{{\rm{x}}^2}{y^4}}}\\\dfrac{3}{{x{y^4}}} = \dfrac{{3.2{\rm{x}}}}{{x{y^4}.2{\rm{x}}}} = \dfrac{{6{\rm{x}}}}{{2{{\rm{x}}^2}{y^4}}}\end{array}\)
b) Ta có:
\(\begin{array}{l}\dfrac{3}{{2{{\rm{x}}^2} - 10{\rm{x}}}} = \dfrac{3}{{2{\rm{x}}\left( {x - 5} \right)}}\\\dfrac{2}{{{x^2} - 25}} = \dfrac{2}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\end{array}\)
Chọn MTC là: \(2{\rm{x}}\left( {x - 5} \right)\left( {x + 5} \right)\)
Nhân tử phụ của các mẫu thức trên lần lượt là: \(\left( {x + 5} \right);2{\rm{x}}\)
Vậy:
\(\begin{array}{l}\dfrac{3}{{2{{\rm{x}}^2} - 10{\rm{x}}}} = \dfrac{3}{{2{\rm{x}}\left( {x - 5} \right)}} = \dfrac{{3\left( {x + 5} \right)}}{{2{\rm{x}}.\left( {x - 5} \right)\left( {x + 5} \right)}}\\\dfrac{2}{{{x^2} - 25}} = \dfrac{2}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{2.2{\rm{x}}}}{{2{\rm{x}}\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{4{\rm{x}}}}{{2{\rm{x}}\left( {x - 5} \right)\left( {x + 5} \right)}}\end{array}\)
a) Ta có:
\(\begin{array}{l}3{\rm{x}}.10y = 30{\rm{xy}}\\{\rm{2}}{\rm{.15x}}y = 30{\rm{x}}y\end{array}\)
Suy ra: \(3{\rm{x}}.10 = 2.15{\rm{x}}y\) nên \(\dfrac{{3{\rm{x}}}}{2} = \dfrac{{15{\rm{x}}y}}{{10y}}\)
b) Ta có:
\(\begin{array}{l}\left( {3{\rm{x}} - 3y} \right).2 = 2.3\left( {x - y} \right) = 6\left( {x - y} \right)\\\left( { - 3} \right).\left( {2y - 2{\rm{x}}} \right) = \left( { - 3} \right).\left( { - 2} \right)\left( {x - y} \right) = 6\left( {x - y} \right)\end{array}\)
Suy ra: \(2.\left( {3{\rm{x}} - 3y} \right) = \left( { - 3} \right).\left( {2y - 2{\rm{x}}} \right)\) nên \(\dfrac{{3{\rm{x}} - 3y}}{{2y - 2{\rm{x}}}} = \dfrac{{ - 3}}{2}\)
c) Ta có: \(\begin{array}{l}\left( {{x^2} - x + 1} \right).x\left( {x + 1} \right) = x.\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) = x.\left( {{x^3} + 1} \right)\\x.\left( {{x^3} + 1} \right)\end{array}\)
Suy ra: \(\left( {{x^2} - x + 1} \right).x.\left( {x + 1} \right) = x.\left( {{x^3} + 1} \right)\) nên \(\dfrac{{{x^2} - x + 1}}{x} = \dfrac{{{x^3} + 1}}{{x\left( {x + 1} \right)}}\)
a) Chọn MTC là: \(\left( {x - 3y} \right)\left( {x + 3y} \right)\)
Nhân tử phụ của các mẫu thức \(\dfrac{2}{{x - 3y}}\) và \(\dfrac{3}{{x + 3y}}\) lần lượt là: \(\left( {x + 3y} \right);\left( {x - 3y} \right)\)
Vậy:
\(\dfrac{2}{{x - 3y}} = \dfrac{{2\left( {x + 3y} \right)}}{{\left( {x - 3y} \right)\left( {x + 3y} \right)}}\)
\(\dfrac{3}{{x + 3y}} = \dfrac{{3.\left( {x - 3y} \right)}}{{\left( {x + 3y} \right)\left( {x - 3y} \right)}}\)
b) Ta có: \(\begin{array}{l}4{\rm{x}} + 24 = 4\left( {x + 6} \right)\\{x^2} - 36 = \left( {x - 6} \right)\left( {x + 6} \right)\end{array}\)
Chọn MTC là: \(4\left( {x + 6} \right)\left( {x - 6} \right)\)
Nhân tử phụ của các phân thức \(\dfrac{7}{{4{\rm{x}} + 24}}\) và \(\dfrac{{13}}{{{x^2} - 36}}\) lần lượt là \(\left( {x - 6} \right);4\)
Vậy:
\(\dfrac{7}{{4{\rm{x}} + 24}} = \dfrac{7}{{4\left( {x + 6} \right)}} = \dfrac{{7\left( {x - 6} \right)}}{{4\left( {x + 6} \right)\left( {x - 6} \right)}}\)
\(\dfrac{{13}}{{{x^2} - 36}} = \dfrac{{13}}{{\left( {x + 6} \right)\left( {x - 6} \right)}} = \dfrac{{13.4}}{{4\left( {x + 6} \right)\left( {x - 6} \right)}} = \dfrac{{52}}{{4\left( {x + 6} \right)\left( {x - 6} \right)}}\)
Ta có:
\(\begin{array}{l}\dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{4{{\rm{x}}^2} - 9{y^2}}} - \dfrac{{x + 8y}}{{3y - 2{\rm{x}}}}\\ = \dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{{{\left( {2{\rm{x}}} \right)}^2} - {{\left( {3y} \right)}^2}}} + \left( { - \dfrac{{x + 8y}}{{3y - 2{\rm{x}}}}} \right)\\ = \dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} + \dfrac{{x + 8y}}{{2{\rm{x}} - 3y}}\\ = \dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} + \dfrac{{x + 8y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{2{\rm{x}} + 3y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{{{\left( {2{\rm{x}} + 3y} \right)}^2}}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{4{{\rm{x}}^2} + 12{\rm{x}}y + 9{y^2} - 24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2}}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} = \dfrac{{{{\left( {2{\rm{x}} - 3y} \right)}^2}}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} = \dfrac{{2{\rm{x}} - 3y}}{{2{\rm{x}} + 3y}}\end{array}\)