a,x mũ 2 y mũ4 -2xymũ2+1 b,a.b+1/4+a mũ 2 bmũ2 c,4+4x+xmũ2 d,x mũ 2 y mũ2+2xy+1 bài 2: a,x mũ 2 +2x+1+(x-1) mũ 2 b,(x+y) mũ 2/x+y c,(x+y) mũ 2+(x-y) m

0 bình luận về “a,x mũ 2 y mũ4 -2xymũ2+1 b,a.b+1/4+a mũ 2 bmũ2 c,4+4x+xmũ2 d,x mũ 2 y mũ2+2xy+1 bài 2: a,x mũ 2 +2x+1+(x-1) mũ 2 b,(x+y) mũ 2/x+y c,(x+y) mũ 2+(x-y) m”

1. Đáp án:

   $\begin{array}{l}
   a){x^2}{y^4} – 2x{y^2} + 1 = {\left( {x{y^2} – 1} \right)^2}\\
   b)\frac{{ab + 1}}{{4 + {a^2}{b^2}}}\\
   c)4 + 4x + {x^2}\\
    = {\left( {x + 2} \right)^2}\\
   d){x^2}{y^2} + 2xy + 1\\
    = {\left( {xy + 1} \right)^2}\\
   B2)a){x^2} + 2x + 1 + {\left( {x – 1} \right)^2}\\
    = {x^2} + 2x + 1 + {x^2} – 2x + 1\\
    = 2{x^2} + 2\\
    = 2\left( {{x^2} + 1} \right)\\
   b)\frac{{{{\left( {x + y} \right)}^2}}}{{x + y}} = x + y\\
   c)\frac{{{{\left( {x + y} \right)}^2} + {{\left( {x – y} \right)}^2} + 2\left( {x – y} \right)\left( {x + y} \right)}}{{2{x^2}}}\\
    = \frac{{{{\left( {x + y + x – y} \right)}^2}}}{{2{x^2}}}\\
    = \frac{{{{\left( {2x} \right)}^2}}}{{2{x^2}}}\\
    = \frac{{4{x^2}}}{{2{x^2}}}\\
    = 2\\
   d){\left( {x.y} \right)^3} – {\left( {x + y} \right)^3}\\
    = \left( {x.y – x – y} \right)\left( {{x^2}{y^2} + xy\left( {x + y} \right) + {{\left( {x + y} \right)}^2}} \right)\\
    = \left( {xy – x – y} \right)\left( {{x^2}{y^2} + {x^2}y + x{y^2} + {x^2} + 2xy + {y^2}} \right)
   \end{array}$

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