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a)
\(\begin{array}{l}A = \left( {\dfrac{{x + 1}}{{2{\rm{x}} - 2}} + \dfrac{3}{{{x^2} - 1}} - \dfrac{{x + 3}}{{2{\rm{x}} + 2}}} \right).\dfrac{{4{{\rm{x}}^2} - 4}}{5}\\A = \left[ {\dfrac{{x + 1}}{{2\left( {x - 1} \right)}} + \dfrac{3}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{x + 3}}{{2\left( {x + 1} \right)}}} \right].\dfrac{{4\left( {{x^2} - 1} \right)}}{5}\end{array}\)
Điều kiện xác định của biểu thức A là: \(x + 1 \ne 0;x - 1 \ne 0\)
b)
\(\begin{array}{l}A = \left( {\dfrac{{x + 1}}{{2{\rm{x}} - 2}} + \dfrac{3}{{{x^2} - 1}} - \dfrac{{x + 3}}{{2{\rm{x}} + 2}}} \right).\dfrac{{4{{\rm{x}}^2} - 4}}{5}\\A = \left[ {\dfrac{{x + 1}}{{2\left( {x - 1} \right)}} + \dfrac{3}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{x + 3}}{{2\left( {x + 1} \right)}}} \right].\dfrac{{4\left( {{x^2} - 1} \right)}}{5}\\A = \dfrac{{\left( {x + 1} \right)\left( {x + 1} \right) - 3.2 - \left( {x + 3} \right)\left( {x - 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{5}\\A = \dfrac{{{x^2} + 2{\rm{x}} + 1 - 6 - {x^2} - 2{\rm{x + 3}}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{5}\\A = \dfrac{{10.4}}{{2.5}} = 4\end{array}\)
Vậy giá trị của A = 4 không phụ thuộc vào các giá trị của biến
\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right).2\left( {x - 2} \right)}}{{4.\left( {x - 2} \right).\left( {x + 2} \right)}} = \dfrac{3}{2}\)
\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}} = \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)\left( {x + 5} \right)}}{{2\left( {x + 5} \right).\left( { - 1} \right)\left( {x - 6} \right)}} = \dfrac{{x + 6}}{{ - 2}} = \dfrac{{-x- 6}}{{ 2}}\)
\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}} = \dfrac{{\left( {1 - y} \right)\left( {1 + y + {y^2}} \right).5\left( {y + 1} \right)}}{{\left( {y + 1} \right).\left( {{y^2} + y + 1} \right)}} = 5\left( {1 - y} \right)\)
\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right) = \dfrac{{\left( {x + 2y} \right).\left( {2{\rm{x}} - y} \right)}}{{{{\left( {2{\rm{x}} - y} \right)}^2}}} = \dfrac{{x + 2y}}{{2{\rm{x}} - y}}\)
a)
\(\begin{array}{l}A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\\A = x - 0,2 - \dfrac{1}{3}x - 2 + 2 - \dfrac{2}{3}x\\ = \left( {x - \dfrac{1}{3}x - \dfrac{2}{3}x} \right) + \left( {\dfrac{{ - 1}}{2} - 2 + 2} \right)\\ = - \dfrac{1}{2}\end{array}\)
Vậy \(A = - \dfrac{1}{2}\) không phụ thuộc vào biến x
b)
\(\begin{array}{l}B = \left( {x - 2y} \right)\left( {{x^2} + 2{\rm{x}}y + 4{y^2}} \right) - \left( {{x^3} - 8{y^3} + 10} \right)\\B = \left[ {x - {{\left( {2y} \right)}^3}} \right] - {x^3} + 8{y^3} - 10\\B = {x^3} - 8{y^3} - {x^3} + 8{y^3} - 10 = - 10\end{array}\)
Vậy B = -10 không phụ thuộc vào biến x, y.
c)
\(\begin{array}{l}C = 4{\left( {x + 1} \right)^2} + {\left( {2{\rm{x}} - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) - 4{\rm{x}}\\{\rm{C = 4}}\left( {{x^2} + 2{\rm{x}} + 1} \right) + \left( {4{{\rm{x}}^2} - 4{\rm{x}} + 1} \right) - 8\left( {{x^2} - 1} \right) - 4{\rm{x}}\\C = 4{{\rm{x}}^2} + 8{\rm{x}} + 4 + 4{{\rm{x}}^2} - 4{\rm{x}} + 1 - 8{{\rm{x}}^2} + 8 - 4{\rm{x}}\\C = \left( {4{{\rm{x}}^2} + 4{{\rm{x}}^2} - 8{{\rm{x}}^2}} \right) + \left( {8{\rm{x}} - 4{\rm{x}} - 4{\rm{x}}} \right) + \left( {4 + 1 + 8} \right)\\C = 13\end{array}\)
Vậy C = 13 không phụ thuộc vào biến x
a) Các biểu thức: \(\dfrac{1}{5}x{y^2}{z^3}; - \dfrac{3}{2}{x^4}{\rm{yx}}{{\rm{z}}^2}\) là đơn thức
b) Các biểu thức: \(2 - x + y; - 5{{\rm{x}}^2}y{z^3} + \dfrac{1}{3}x{y^2}z + x + 1\) là đa thức
\(\begin{array}{l}a)\dfrac{x}{{xy + {y^2}}} - \dfrac{y}{{{x^2} + xy}}\\ = \dfrac{x}{{y\left( {x + y} \right)}} - \dfrac{y}{{x\left( {x + y} \right)}}\\ = \dfrac{{{x^2} - {y^2}}}{{xy\left( {x + y} \right)}} = \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{xy\left( {x + y} \right)}} = \dfrac{{x - y}}{{xy}}\end{array}\)
\(\begin{array}{l}b)\dfrac{{{x^2} + 4}}{{{x^2} - 4}} - \dfrac{x}{{x + 2}} - \dfrac{x}{{2 - x}}\\ = \dfrac{{{x^2} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{x}{{x + 2}} + \dfrac{x}{{x - 2}}\\ = \dfrac{{{x^2} + 4 - x\left( {x - 2} \right) + x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{{x^2} + 4 - {x^2} + 2{\rm{x}} + {x^2} + 2{\rm{x}}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} + 4{\rm{x}} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{x + 2}}{{x - 2}}\end{array}\)
\(\begin{array}{l}c)\dfrac{{{a^2} + ab}}{{b - a}}:\dfrac{{a + b}}{{2{{\rm{a}}^2} - 2{b^2}}}\\ = \dfrac{{a\left( {a + b} \right)}}{{b - a}}.\dfrac{{2{{\rm{a}}^2} - 2{b^2}}}{{a + b}}\\ = \dfrac{{a\left( {a + b} \right).2.\left( {{a^2} - {b^2}} \right)}}{{ - \left( {a - b} \right).\left( {a + b} \right)}}\\ = \dfrac{{a\left( {a + b} \right).2.\left( {a - b} \right).\left( {a + b} \right)}}{{ - \left( {a - b} \right)\left( {a + b} \right)}} = - 2{\rm{a}}\left( {a + b} \right)\end{array}\)
\(\begin{array}{l}d)\left( {\dfrac{{2{\rm{x}} + 1}}{{2{\rm{x}} - 1}} - \dfrac{{2{\rm{x}} - 1}}{{2{\rm{x}} + 1}}} \right):\dfrac{{4{\rm{x}}}}{{10{\rm{x}} - 5}}\\ = \dfrac{{{{\left( {2{\rm{x}} + 1} \right)}^2} - {{\left( {2{\rm{x}} - 1} \right)}^2}}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} - 1} \right)}}.\dfrac{{10x - 5}}{{4{\rm{x}}}}\\ = \dfrac{{\left( {2{\rm{x}} + 1 + 2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1 - 2{\rm{x}} + 1} \right)}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} - 1} \right)}}.\dfrac{{5.\left( {2{\rm{x}} - 1} \right)}}{{4{\rm{x}}}}\\ = \dfrac{{4{\rm{x}}.2.5\left( {2{\rm{x}} - 1} \right)}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} - 1} \right).4{\rm{x}}}} = \dfrac{{10}}{{2{\rm{x}} + 1}}\end{array}\)
a) Vì x = 1,2 và x + y = 6,2 nên \(y = 6,2 - x = 6,2 - 1,2 = 5\)
\(\begin{array}{l}P = \left( {5{{\rm{x}}^2} - 2{\rm{x}}y + {y^2}} \right) - \left( {{x^2} + {y^2}} \right) - \left( {4{{\rm{x}}^2} - 5{\rm{x}}y + 1} \right)\\P = 5{{\rm{x}}^2} - 2{\rm{x}}y + {y^2} - {x^2} - {y^2} - 4{{\rm{x}}^2} + 5{\rm{x}}y - 1\\P = \left( {5{{\rm{x}}^2} - {x^2} - 4{{\rm{x}}^2}} \right) + \left( {{y^2} - {y^2}} \right) + \left( { - 2{\rm{x}}y + 5{\rm{x}}y} \right)\\P = 3{\rm{x}}y - 1 \end{array}\)
Thay x = 1,2; y = 5 vào biểu thức P = 3xy - 1 ta được
\(P = 3.1,2.5 - 1 = 17\)
Vậy P = 17
b) Ta có:
\(\begin{array}{l}\left( {{x^2} - 5{\rm{x}} + 4} \right)\left( {2{\rm{x}} + 3} \right) - \left( {2{{\rm{x}}^2} - x - 10} \right)\left( {x - 3} \right)\\ = {x^2}.2{\rm{x}} + {x^2}.3 - 5{\rm{x}}.2{\rm{x}} - 5{\rm{x}}.3 + 4.2{\rm{x}} + 4.3 - {\rm{[2}}{{\rm{x}}^2}.x + 2{{\rm{x}}^2}.( - 3) - x.x - x.( - 3) - 10.x - 10.( - 3){\rm{]}}\\ = 2{{\rm{x}}^3} + 3{{\rm{x}}^2} - 10{{\rm{x}}^2} - 15{\rm{x}} + 8{\rm{x}} + 12 - 2{{\rm{x}}^3} + 6{\rm{x}}{}^2 + {x^2} - 3{\rm{x}} + 10{\rm{x}} - 30\\ = \left( {2{{\rm{x}}^3} - 2{{\rm{x}}^3}} \right) + \left( {3{{\rm{x}}^2} - 10{{\rm{x}}^2} + 6{{\rm{x}}^2} + {x^2}} \right) + ( - 15{\rm{x}} + 8{\rm{x}} - 3{\rm{x}} + 10{\rm{x}}) +(12-30)\\ = - 18\end{array}\)
Vậy biểu thức đã cho bằng -18 nên không phụ thuộc vào biến x
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:
\(\begin{array}{l}\dfrac{3}{{2{{\rm{x}}^2} - 10{\rm{x}}}} = \dfrac{3}{{2{\rm{x}}\left( {x - 5} \right)}} = \dfrac{{3\left( {x + 5} \right)}}{{2{\rm{x}}.\left( {x - 5} \right)\left( {x + 5} \right)}}\\\dfrac{2}{{{x^2} - 25}} = \dfrac{2}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{2.2{\rm{x}}}}{{2{\rm{x}}\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{4{\rm{x}}}}{{2{\rm{x}}\left( {x - 5} \right)\left( {x + 5} \right)}}\end{array}\)
\(\begin{array}{l}a)\dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{{x^2} - 4}}{{x + 1}}.\dfrac{{x - 2}}{{y + 6}}\\ = \dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{x - 2}}{{y + 6}}.\dfrac{{{x^2} - 4}}{{x + 1}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 4{\rm{x}} + 4} \right).\left( {y + 6} \right).\left( {x + 1} \right)}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {x - 2} \right)\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}.\left( {y + 6} \right).\left( {x + 1} \right)}} = \dfrac{{x + 2}}{{x + 1}}\end{array}\)
\(\begin{array}{l}b)\left(\frac{2x+1}{{x - 3}} + \frac{2x+1}{x+3}\right ) .\dfrac{{x^2 - 9}}{{2{\rm{x}} + 1}} \\ = (2x+1) \left ( \frac {1}{x-3} + \frac {1}{x+3} \right ) . \frac {(x-3)(x+3)}{2x + 1} \\ = (2x+1) \frac {x+3 + x - 3}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x + 1} \\ = \frac {2x(2x+1)}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x +1} \\= 2x \end{array}\)
\(a)\dfrac{{20{\rm{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{\rm{x}}^2}}}{{6y}}} \right) = \dfrac{{20{\rm{x}}}}{{3{y^2}}}.\left( { - \dfrac{{6y}}{{15{{\rm{x}}^2}}}} \right) = \dfrac{{20{\rm{x}}.\left( { - 6y} \right)}}{{3{y^2}.15{{\rm{x}}^2}}} = \dfrac{{ - 8}}{{3{\rm{x}}y}}\)
\(b)\dfrac{{9{{\rm{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{\rm{x}} + y}}{{2{\rm{x}} + 2y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right)}}{{x + y}}.\dfrac{{2{\rm{x}} + 2y}}{{3{\rm{x}} + y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right).2.\left( {x + y} \right)}}{{(x + y).\left( {3{\rm{x}} + y} \right)}} = 2\left( {3{\rm{x}} - y} \right)\)
\(\begin{array}{l}c)\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}} = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{y - x}}.\dfrac{{{x^2} - 2{\rm{x}}y + {y^2}}}{{{x^2} - xy + {y^2}}}\\ = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right).{{\left( {x - y} \right)}^2}}}{{ - (x - y)\left( {{x^2} - xy + {y^2}} \right)}} = \left( {x + y} \right)\left( {y - x} \right) = {{y^2} - {x^2}} \end{array}\)
\(d)\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right) = \dfrac{{\left( {3 - x} \right)\left( {3 + x} \right)}}{x}.\dfrac{1}{{x - 3}} = \dfrac{{ - \left( {x - 3} \right)\left( {3 + x} \right)}}{{x.\left( {x - 3} \right)}} = \dfrac{{ - \left( {3 + x} \right)}}{x}.\)
a)
\(\begin{array}{l}B = \left( {\dfrac{{5{\rm{x}} + 2}}{{{x^2} - 10{\rm{x}}}} + \dfrac{{5{\rm{x}} - 2}}{{{x^2} + 10{\rm{x}}}}} \right).\dfrac{{{x^2} - 100}}{{{x^2} + 4}}\\B = \left[ {\dfrac{{5{\rm{x}} + 2}}{{x\left( {x - 10} \right)}} + \dfrac{{5{\rm{x - }}2}}{{x\left( {x + 10} \right)}}} \right].\dfrac{{\left( {x - 10} \right)\left( {x + 10} \right)}}{{{x^2} + 4}}\end{array}\)
Điều kiện xác định của biểu thức B là: \(x\left( {x - 10} \right) \ne 0;x\left( {x + 10} \right) \ne 0\) hay \( x \not \in \left\{ {0; -10 ; 10} \right\} \)
b) Ta có:
\(\begin{array}{l}B = \left( {\dfrac{{5{\rm{x}} + 2}}{{{x^2} - 10{\rm{x}}}} + \dfrac{{5{\rm{x}} - 2}}{{{x^2} + 10{\rm{x}}}}} \right).\dfrac{{{x^2} - 100}}{{{x^2} + 4}}\\B = \left[ {\dfrac{{5{\rm{x}} + 2}}{{x\left( {x - 10} \right)}} + \dfrac{{5{\rm{x - }}2}}{{x\left( {x + 10} \right)}}} \right].\dfrac{{\left( {x - 10} \right)\left( {x + 10} \right)}}{{{x^2} + 4}}\\B = \dfrac{{\left( {5{\rm{x}} + 2} \right)\left( {x + 10} \right) + \left( {5{\rm{x}} - 2} \right)\left( {x - 10} \right)}}{{x\left( {x - 10} \right)\left( {x + 10} \right)}}.\dfrac{{\left( {x - 10} \right)\left( {x + 10} \right)}}{{{x^2} + 4}}\\B = \dfrac{{5{{\rm{x}}^2} + 52{\rm{x}} + 20 + 5{{\rm{x}}^2} - 52{\rm{x}} + 20}}{{x\left( {x - 10} \right)\left( {x + 10} \right)}}.\dfrac{{\left( {x - 10} \right)\left( {x + 10} \right)}}{{{x^2} + 4}}\\B = \dfrac{{10\left( {{x^2} + 4} \right).\left( {x - 10} \right)\left( {x + 10} \right)}}{{x\left( {x - 10} \right)\left( {x + 10} \right).\left( {{x^2} + 4} \right)}} = \dfrac{{10}}{x}\end{array}\)
Với x = 0,1 ta có:
\(B = \dfrac{{10}}{{0,1}} = 100\)
c) Để B nguyên thì \(\dfrac{{10}}{x}\) nguyên
Suy ra x \( \in \) Ư (10) = \(\left\{ { \pm 1; \pm 2; \pm 5; \pm 10} \right\}\)
Mà \( x \not \in \left\{ {0; -10 ; 10} \right\} \)
Vậy \(x \in \left\{ { \pm 1; \pm 2; \pm 5} \right\}\) thì B nguyên