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Thay x = -1 , y = 2 vào đa thức P ta được:
\(\begin{array}{l}P = {\left( { - 1} \right)^3}.2 - 14.{2^3} - 6.\left( { - 1} \right).2^2 + 2 + 2\\P = - 2 - 112 + 24 + 4 = -86\end{array}\)
Vậy đa thức P = -86 tại x = -1; y = 2
\(\begin{array}{l}a)3{{\rm{x}}^2} - 6{\rm{x}}y + 3{y^2} - 5{\rm{x}} + 5y\\ = \left( {3{{\rm{x}}^2} - 6{\rm{x}}y + 3{y^2}} \right) - \left( {5{\rm{x}} - 5y} \right)\\ = 3\left( {{x^2} - 2{\rm{x}}y + {y^2}} \right) - 5\left( {x - y} \right)\\ = 3{\left( {x - y} \right)^2} - 5\left( {x - y} \right)\\ = \left( {x - y} \right)\left[ {3\left( {x - y} \right) - 5} \right] = \left( {x - y} \right)\left( {3{\rm{x}} - 3y - 5} \right)\end{array}\)
\(\begin{array}{l}b)2{{\rm{x}}^2}y + 4{\rm{x}}{y^2} + 2{y^3} - 8y\\ = 2y\left[ {\left( {{x^2} + 2{\rm{x}}y + {y^2}} \right) - 4} \right]\\ = 2y\left[ {{{\left( {x + y} \right)}^2} - {2^2}} \right]\\ = 2y\left( {x + y + 2} \right)\left( {x + y - 2} \right)\end{array}\)
\(a)4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2} = {\left( {2{\rm{x}}} \right)^2} - 2.2{\rm{x}}.3y + {\left( {3y} \right)^2} = {\left( {2{\rm{x}} - 3y} \right)^2}\)
\(b){x^3} + 9{{\rm{x}}^2} + 27{\rm{x}} + 27 = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {\left( {x + 3} \right)^3}\)
\(c)8{y^3} - 12{y^2} + 6y - 1 = {\left( {2y} \right)^3} - 3.{\left( {2y} \right)^2}.1 + 3.2y{.1^2} - {1^3} = {\left( {2y - 1} \right)^3}\)
\(\begin{array}{l}d) {\left( {2{\rm{x}} + y} \right)^2} - 4{y^2}\\ = {\left( {2{\rm{x}} + y} \right)^2} - {\left( {2y} \right)^2}\\ = \left( {2{\rm{x}} + y + 2y} \right)\left( {2{\rm{x}} + y - 2y} \right) = \left( {2{\rm{x}} + 3y} \right)\left( {2{\rm{x}} - y} \right)\end{array}\)
\(e) 27{y^3} + 8 = {\left( {3y} \right)^3} + {2^3} = \left( {3y + 2} \right)\left( {9{y^2} - 6y + 4} \right)\)
\(g) 64 - 125{{\rm{x}}^3} = {4^3} - {\left( {5{\rm{x}}} \right)^3} = \left( {4 - 5{\rm{x}}} \right)\left( {16 + 20{\rm{x}} + 25{{\rm{x}}^2}} \right)\)
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
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) Ta có:
\(\begin{array}{l}B - C = \left( {2{{\rm{x}}^2} - {y^2}} \right) - \left( {{x^2} - 3{\rm{x}}y} \right)\\B - C = 2{{\rm{x}}^2} - {y^2} - {x^2} + 3{\rm{x}}y\\B - C = \left( {2{{\rm{x}}^2} - {x^2}} \right) + 3{\rm{x}}y - {y^2} = {x^2} + 3{\rm{x}}y - {y^2}\end{array}\)
b) Ta có:
\(\begin{array}{l}(B - C) + A = {\rm{[}}\left( {2{{\rm{x}}^2} - {y^2}} \right) - \left( {{x^2} - 3{\rm{x}}y} \right){\rm{] + (}}{{\rm{x}}^2} - 2{\rm{x}}y + {y^2})\\(B - C) + A = {x^2} + 3{\rm{x}}y - {y^2} + {x^2} - 2{\rm{x}}y + {y^2}\\(B - C) + A = \left( {{x^2} + {x^2}} \right) + \left( {3{\rm{x}}y - 2{\rm{x}}y} \right) + \left( {{y^2} - {y^2}} \right)\\(B - C) + A = 2{{\rm{x}}^2} + xy\end{array}\)
\(a)\left( { - \frac{{3{\rm{x}}}}{{5{\rm{x}}{y^2}}}} \right).\left( { - \frac{{5{y^2}}}{{12{\rm{x}}y}}} \right) = \frac{{\left( { - 3{\rm{x}}} \right).\left( { - 5{y^2}} \right)}}{{5{\rm{x}}{y^2}.12{\rm{x}}y}} = \frac{1}{{4{\rm{x}}y}}\)
\(b)\frac{{{x^2} - x}}{{2{\rm{x}} + 1}}.\frac{{4{{\rm{x}}^2} - 1}}{{{x^3} - 1}} = \frac{{x\left( {x - 1} \right).\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}{{\left( {2{\rm{x}} + 1} \right).\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{x\left( {2{\rm{x}} - 1} \right)}}{{{x^2} + x + 1}}\)
\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right).2\left( {x - 2} \right)}}{{4.\left( {x - 2} \right).\left( {x + 2} \right)}} = \dfrac{3}{2}\)
\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}} = \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)\left( {x + 5} \right)}}{{2\left( {x + 5} \right).\left( { - 1} \right)\left( {x - 6} \right)}} = \dfrac{{x + 6}}{{ - 2}} = \dfrac{{-x- 6}}{{ 2}}\)
\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}} = \dfrac{{\left( {1 - y} \right)\left( {1 + y + {y^2}} \right).5\left( {y + 1} \right)}}{{\left( {y + 1} \right).\left( {{y^2} + y + 1} \right)}} = 5\left( {1 - y} \right)\)
\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right) = \dfrac{{\left( {x + 2y} \right).\left( {2{\rm{x}} - y} \right)}}{{{{\left( {2{\rm{x}} - y} \right)}^2}}} = \dfrac{{x + 2y}}{{2{\rm{x}} - y}}\)
\(a)\left( { - \frac{{3{\rm{x}}}}{{5{\rm{x}}{y^2}}}} \right):\left( { - \frac{{5{y^2}}}{{12{\rm{x}}y}}} \right) = \frac{{ - 3{\rm{x}}}}{{5{\rm{x}}{y^2}}}.\frac{{ - 12{\rm{x}}y}}{{5{y^2}}} = \frac{{36{{\rm{x}}^2}y}}{{25{\rm{x}}{y^4}}}\)
b) \(\frac{4{{\text{x}}^{2}}-1}{8{{\text{x}}^{3}}-1}:\frac{4{{\text{x}}^{2}}+4\text{x}+1}{4{{\text{x}}^{2}}+2\text{x}+1}=\frac{4{{\text{x}}^{2}}-1}{8{{\text{x}}^{3}}-1}.\frac{4{{\text{x}}^{2}}+2\text{x}+1}{4{{\text{x}}^{2}}+4\text{x}+1}\)
\(=\frac{\left( 2\text{x}-1 \right)\left( 2\text{x}+1 \right)\left( 4{{\text{x}}^{2}}+2\text{x}+1 \right)}{\left( 2\text{x}-1 \right)\left( 4{{\text{x}}^{2}}+2\text{x}+1 \right){{\left( 2\text{x}+1 \right)}^{2}}}=\frac{1}{2\text{x}+1}\).
a) Thay x = -1, y = 1 vào đa thức A ta được:
\(\begin{array}{l}A = 4.{\left( { - 1} \right)^6} - 2.{\left( { - 1} \right)^2}{.1^3} - 5.\left( { - 1} \right).1 + 2\\A = 4 - 2 + 5 + 2 = 9\end{array}\)
Vậy A =9 tại x = -1; y = 1
Thay x = -1, y = 1 vào đa thức B ta được:
\(\begin{array}{l}B = 3.{\left( { - 1} \right)^2}{.1^3} + 5.\left( { - 1} \right).1 - 7\\B = 3 - 5 - 7 = - 9\end{array}\)
Vậy B = -9 tại x = -1; y = 1
b) Ta có:
\(\begin{array}{l}A + B = \left( {4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2} \right) + \left( {3{{\rm{x}}^2}{y^3} + 5{\rm{x}}y - 7} \right)\\ = 4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2 + 3{{\rm{x}}^2}{y^3} + 5{\rm{x}}y - 7\\ = 4{{\rm{x}}^6} + \left( { - 2{{\rm{x}}^2}{y^3} + 3{{\rm{x}}^2}{y^3}} \right) + \left( { - 5{\rm{x}}y + 5{\rm{x}}y} \right) + 2 - 7\\ = 4{{\rm{x}}^6} + {x^2}{y^3} - 5\end{array}\)
\(\begin{array}{l}A - B = \left( {4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2} \right) - \left( {3{{\rm{x}}^2}{y^3} + 5{\rm{x}}y - 7} \right)\\ = 4{{\rm{x}}^6} - 2{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 2 - 3{{\rm{x}}^2}{y^3} - 5{\rm{x}}y + 7\\ = 4{{\rm{x}}^6} + \left( { - 2{{\rm{x}}^2}{y^3} - 3{{\rm{x}}^2}{y^3}} \right) + \left( { - 5{\rm{x}}y - 5{\rm{x}}y} \right) + 2 + 7\\ = 4{{\rm{x}}^6} - 5{x^2}{y^3} - 10{\rm{x}}y + 9\end{array}\)