a) $x^{2}$ + 5x – $y^{2}$ – 5y b) $2x^{2}$ + 4xy + x + 2y c) xy + x + $x^{2}$y + 1 d) $x^{2}$ – 4 + ($x-2)^{2}$ e) $x^{2}$ – 4x – $4y^{2}$ + 4

a) $x^{2}$ + 5x – $y^{2}$ – 5y
b) $2x^{2}$ + 4xy + x + 2y
c) xy + x + $x^{2}$y + 1
d) $x^{2}$ – 4 + ($x-2)^{2}$
e) $x^{2}$ – 4x – $4y^{2}$ + 4
f) 45+ $x^{3}$ – $5x^{2}$ – 9x
g) $x^{3}$ – $3x^{2}$ – 4x + 12
h) $x^{3}$ – $3x^{2}$ + 6x – 18
i) $x^{3}$ – $4x^{2}$ – 12x + 27
PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ BẰNG CÁCH PHỐI HỢP NHIỀU PHƯƠNG PHÁP

0 bình luận về “a) $x^{2}$ + 5x – $y^{2}$ – 5y b) $2x^{2}$ + 4xy + x + 2y c) xy + x + $x^{2}$y + 1 d) $x^{2}$ – 4 + ($x-2)^{2}$ e) $x^{2}$ – 4x – $4y^{2}$ + 4”

  1. Đáp án:

    $\begin{array}{l}
    a){x^2} + 5x – {y^2} – 5y\\
     = {x^2} – {y^2} + 5x – 5y\\
     = \left( {x – y} \right)\left( {x + y} \right) + 5\left( {x – y} \right)\\
     = \left( {x – y} \right)\left( {x + y + 5} \right)\\
    b)2{x^2} + 4xy + x + 2y\\
     = 2x\left( {x + 2y} \right) + x + 2y\\
     = \left( {x + 2y} \right)\left( {2x + 1} \right)\\
    c)xy + x + {x^2}y + 1\\
     = xy + {x^2}y + x + 1\\
     = xy\left( {x + 1} \right) + x + 1\\
     = \left( {x + 1} \right)\left( {xy + 1} \right)\\
    d){x^2} – 4 + {\left( {x – 2} \right)^2}\\
     = \left( {x – 2} \right)\left( {x + 2} \right) + {\left( {x – 2} \right)^2}\\
     = \left( {x – 2} \right)\left( {x + 2 + x – 2} \right)\\
     = \left( {x – 2} \right).2x\\
    e){x^2} – 4x – 4{y^2} + 4\\
     = {x^2} – 4x + 4 – 4{y^2}\\
     = {\left( {x – 2} \right)^2} – {\left( {2y} \right)^2}\\
     = \left( {x – 2 – 2y} \right)\left( {x – 2 + 2y} \right)\\
    f)45 + {x^3} – 5{x^2} – 9x\\
     = {x^2}\left( {x – 5} \right) – 9\left( {x – 5} \right)\\
     = \left( {x – 5} \right)\left( {{x^2} – 9} \right)\\
     = \left( {x – 5} \right)\left( {x + 3} \right)\left( {x – 3} \right)\\
    g){x^3} – 3{x^2} – 4x + 12\\
     = {x^2}\left( {x – 3} \right) – 4\left( {x – 3} \right)\\
     = \left( {x – 3} \right)\left( {{x^2} – 4} \right)\\
     = \left( {x – 3} \right)\left( {x – 2} \right)\left( {x + 2} \right)\\
    h){x^3} – 3{x^2} + 6x – 18\\
     = {x^2}\left( {x – 3} \right) + 6\left( {x – 3} \right)\\
     = \left( {x – 3} \right)\left( {{x^2} + 6} \right)\\
    i){x^3} – 4{x^2} – 12x + 27\\
     = {x^3} + 3{x^2} – 7{x^2} – 21x + 9x + 27\\
     = {x^2}\left( {x + 3} \right) – 7x\left( {x + 3} \right) + 9\left( {x + 3} \right)\\
     = \left( {x + 3} \right)\left( {{x^2} – 7x + 9} \right)
    \end{array}$

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