Câu 1 và 3 giúp em ạ

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Câu 1 và 3 giúp em ạ
cau-1-va-3-giup-em-a

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King 6 years 2020-11-28T05:18:41+00:00 1 Answers 56 views 0

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  1. Amity
    0
    2020-11-28T05:20:12+00:00
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    Đáp án:

    $\begin{array}{l}
    1){\left( {x + 2} \right)^2} = {x^2} + 2.x.2 + {2^2}\\
     = {x^2} + 4x + 4\\
    2){\left( {z – 1} \right)^2} = {z^2} – 2z + 1\\
    3){\left( {2t + 3} \right)^2} = 4{t^2} + 12t + 9\\
    4){\left( {3a – 1} \right)^2} = 9{a^2} – 6a + 1\\
    5){\left( {4 + U} \right)^3} = {4^3} + {3.4^2}.U + 3.4.{U^2} + {U^3}\\
     = {U^3} + 12{U^2} + 48U + 64\\
    6){\left( {k – 3} \right)^3} = {k^3} – 9{k^2} + 27k – 27\\
    7){\left( {2m + 1} \right)^3} = 8{m^3} + 12{m^2} + 6m + 1\\
    8){\left( {1 – 3b} \right)^3}\\
     = 1 – 3.1.3b + 3.1.9{b^2} – 27{b^3}\\
     = 1 – 9b + 27{b^2} – 27{b^3}\\
    B3)\\
    1)15{x^2} – 5x – {\left( {2x – 3} \right)^2}\\
     = 15{x^2} – 5x – \left( {4{x^2} – 12x + 9} \right)\\
     = 15{x^2} – 5x – 4{x^2} + 12x – 9\\
     = 11{x^2} + 7x – 9\\
    2)\\
    {\left( {3 + x} \right)^2} + {\left( {x – 2} \right)^2}\\
     = {x^2} + 6x + 9 + {x^2} – 4x + 4\\
     = 2{x^2} + 2x + 13\\
    3){\left( {2x + 3} \right)^2} – \left( {3x – 1} \right)\left( {x + 4} \right)\\
     = 4{x^2} + 12x + 9 – \left( {3{x^2} + 12x – x – 4} \right)\\
     = {x^2} + x + 13\\
    4){\left( {1 + x} \right)^3} – 3x\left( {2x + 1} \right)\left( {2x – 1} \right)\\
     = {x^3} + 3{x^2} + 3x + 1 – 3x\left( {4{x^2} – 1} \right)\\
     = {x^3} + 3{x^2} + 3x + 1 – 12{x^3} + 3x\\
     =  – 11{x^3} + 3{x^2} + 6x + 1\\
    5){\left( {x – 4} \right)^2} + \left( {x – 3} \right)\left( {x + 3} \right)\\
     = {x^2} – 8x + 16 + {x^2} – 9\\
     = 2{x^2} – 8x + 7\\
    6)\\
    \left( {2x + 1} \right)\left( {4{x^2} – 2x + 1} \right) – {\left( {2x – 1} \right)^3}\\
     = {\left( {2x} \right)^3} + {1^3} – \left( {8{x^3} – 12{x^2} + 6x – 1} \right)\\
     = 8{x^3} + 1 – 8{x^3} + 12{x^2} – 6x + 1\\
     = 12{x^2} – 6x + 2\\
    7){\left( {2x – 1} \right)^2} – \left( {4x – 1} \right)\left( {x + 2} \right)\\
     = 4{x^2} – 4x + 1 – \left( {4{x^2} + 8x – x – 2} \right)\\
     =  – 11x + 3\\
    8){\left( {x + 5} \right)^2} + \left( {x – 1} \right)\left( {4 – x} \right)\\
     = {x^2} + 10x + 25 – {x^2} + 4x – 4 + x\\
     = 15x + 21
    \end{array}$

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