Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(8{{\rm{x}}^3} - 36{{\rm{x}}^2}y + 54{\rm{x}}{y^2} - 27{y^3} = {\left( {2{\rm{x}}} \right)^3} - 3.\left( {2{\rm{x}}} \right).3y + 3.2{\rm{x}}.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {2{\rm{x}} - 3y} \right)^3}\)
a) \({a^3} + 12{{\rm{a}}^2} + 48{\rm{a}} + 64 \\= {a^3} + 3{{\rm{a}}^2}.4 + 3{\rm{a}}{.4^2} + {4^3} \\= {\left( {a + 4} \right)^3}\)
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) \(8{{\rm{a}}^3} - 12{{\rm{a}}^2}b + 6{\rm{a}}{b^2} - {b^3} \\= {\left( {2{\rm{a}}} \right)^2} - 3.{\left( {2{\rm{a}}} \right)^2}.b + 3.2{\rm{a}}.{b^2} - {b^3} \\= {\left( {2{\rm{a}} - b} \right)^3}\)
d) \(27{{\rm{x}}^3} + 54{{\rm{x}}^2}y + 36{\rm{x}}{y^2} + 8{y^3}\\= {\left( {3{\rm{x}}} \right)^3} + 3.{\left( {3{\rm{x}}} \right)^2}.2y + 3.3{\rm{x}}.{\left( {2y} \right)^2} + {\left( {2y} \right)^3} \\= {\left( {3{\rm{x}} + 2y} \right)^3}\)
\(a)4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2} = {\left( {2{\rm{x}}} \right)^2} - 2.2{\rm{x}}.3y + {\left( {3y} \right)^2} = {\left( {2{\rm{x}} - 3y} \right)^2}\)
\(b){x^3} + 9{{\rm{x}}^2} + 27{\rm{x}} + 27 = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {\left( {x + 3} \right)^3}\)
\(c)8{y^3} - 12{y^2} + 6y - 1 = {\left( {2y} \right)^3} - 3.{\left( {2y} \right)^2}.1 + 3.2y{.1^2} - {1^3} = {\left( {2y - 1} \right)^3}\)
\(\begin{array}{l}d) {\left( {2{\rm{x}} + y} \right)^2} - 4{y^2}\\ = {\left( {2{\rm{x}} + y} \right)^2} - {\left( {2y} \right)^2}\\ = \left( {2{\rm{x}} + y + 2y} \right)\left( {2{\rm{x}} + y - 2y} \right) = \left( {2{\rm{x}} + 3y} \right)\left( {2{\rm{x}} - y} \right)\end{array}\)
\(e) 27{y^3} + 8 = {\left( {3y} \right)^3} + {2^3} = \left( {3y + 2} \right)\left( {9{y^2} - 6y + 4} \right)\)
\(g) 64 - 125{{\rm{x}}^3} = {4^3} - {\left( {5{\rm{x}}} \right)^3} = \left( {4 - 5{\rm{x}}} \right)\left( {16 + 20{\rm{x}} + 25{{\rm{x}}^2}} \right)\)
\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right).2\left( {x - 2} \right)}}{{4.\left( {x - 2} \right).\left( {x + 2} \right)}} = \dfrac{3}{2}\)
\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}} = \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)\left( {x + 5} \right)}}{{2\left( {x + 5} \right).\left( { - 1} \right)\left( {x - 6} \right)}} = \dfrac{{x + 6}}{{ - 2}} = \dfrac{{-x- 6}}{{ 2}}\)
\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}} = \dfrac{{\left( {1 - y} \right)\left( {1 + y + {y^2}} \right).5\left( {y + 1} \right)}}{{\left( {y + 1} \right).\left( {{y^2} + y + 1} \right)}} = 5\left( {1 - y} \right)\)
\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right) = \dfrac{{\left( {x + 2y} \right).\left( {2{\rm{x}} - y} \right)}}{{{{\left( {2{\rm{x}} - y} \right)}^2}}} = \dfrac{{x + 2y}}{{2{\rm{x}} - y}}\)
a) 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
\(a)\dfrac{{20{\rm{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{\rm{x}}^2}}}{{6y}}} \right) = \dfrac{{20{\rm{x}}}}{{3{y^2}}}.\left( { - \dfrac{{6y}}{{15{{\rm{x}}^2}}}} \right) = \dfrac{{20{\rm{x}}.\left( { - 6y} \right)}}{{3{y^2}.15{{\rm{x}}^2}}} = \dfrac{{ - 8}}{{3{\rm{x}}y}}\)
\(b)\dfrac{{9{{\rm{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{\rm{x}} + y}}{{2{\rm{x}} + 2y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right)}}{{x + y}}.\dfrac{{2{\rm{x}} + 2y}}{{3{\rm{x}} + y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right).2.\left( {x + y} \right)}}{{(x + y).\left( {3{\rm{x}} + y} \right)}} = 2\left( {3{\rm{x}} - y} \right)\)
\(\begin{array}{l}c)\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}} = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{y - x}}.\dfrac{{{x^2} - 2{\rm{x}}y + {y^2}}}{{{x^2} - xy + {y^2}}}\\ = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right).{{\left( {x - y} \right)}^2}}}{{ - (x - y)\left( {{x^2} - xy + {y^2}} \right)}} = \left( {x + y} \right)\left( {y - x} \right) = {{y^2} - {x^2}} \end{array}\)
\(d)\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right) = \dfrac{{\left( {3 - x} \right)\left( {3 + x} \right)}}{x}.\dfrac{1}{{x - 3}} = \dfrac{{ - \left( {x - 3} \right)\left( {3 + x} \right)}}{{x.\left( {x - 3} \right)}} = \dfrac{{ - \left( {3 + x} \right)}}{x}.\)
a) 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)
\(\begin{array}{l}A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\\A = x - 0,2 - \dfrac{1}{3}x - 2 + 2 - \dfrac{2}{3}x\\ = \left( {x - \dfrac{1}{3}x - \dfrac{2}{3}x} \right) + \left( {\dfrac{{ - 1}}{2} - 2 + 2} \right)\\ = - \dfrac{1}{2}\end{array}\)
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
\(\begin{array}{l}a)\dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{{x^2} - 4}}{{x + 1}}.\dfrac{{x - 2}}{{y + 6}}\\ = \dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{x - 2}}{{y + 6}}.\dfrac{{{x^2} - 4}}{{x + 1}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 4{\rm{x}} + 4} \right).\left( {y + 6} \right).\left( {x + 1} \right)}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {x - 2} \right)\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}.\left( {y + 6} \right).\left( {x + 1} \right)}} = \dfrac{{x + 2}}{{x + 1}}\end{array}\)
\(\begin{array}{l}b)\left(\frac{2x+1}{{x - 3}} + \frac{2x+1}{x+3}\right ) .\dfrac{{x^2 - 9}}{{2{\rm{x}} + 1}} \\ = (2x+1) \left ( \frac {1}{x-3} + \frac {1}{x+3} \right ) . \frac {(x-3)(x+3)}{2x + 1} \\ = (2x+1) \frac {x+3 + x - 3}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x + 1} \\ = \frac {2x(2x+1)}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x +1} \\= 2x \end{array}\)
\(a){x^2} + \dfrac{1}{2}x + \dfrac{1}{{16}} \\= {x^2} + 2.x.\dfrac{1}{4} + {\left( {\dfrac{1}{4}} \right)^2} \\= {\left( {x + \dfrac{1}{4}} \right)^2}\)
\(b)25{{\rm{x}}^2} - 10{\rm{x}}y + {y^2} \\= {\left( {5{\rm{x}}} \right)^2} - 2.5{\rm{x}}.y + {y^2} \\= {\left( {5{\rm{x}} - y} \right)^2}\)
\(\begin{array}{l}c){x^3} + 9{{\rm{x}}^2}y + 27{\rm{x}}{y^2} + 27{y^3}\\ = {x^3} + 3{{\rm{x}}^2}.3y + 3.x.{\left( {3y} \right)^2} + {\left( {3y} \right)^3}\\ = {\left( {x + 3y} \right)^3}\end{array}\)
\(\begin{array}{l}d)64{{\rm{x}}^3} - 48{{\rm{x}}^2}y + 12{\rm{x}}{y^2} - {y^3}\\ = {\left( {4{\rm{x}}} \right)^3} - 3.{\left( {4{\rm{x}}} \right)^2}.y + 3.4{\rm{x}}.{y^2} - {y^3}\\ = {\left( {4{\rm{x}} - y} \right)^3}\end{array}\)