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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: phân tích đa thức thành nhân tử 9x^2 – 16y^2 – 6x + 1 x^2+5x-24 xy-4a^2xy 3x^2 – 5x – 2 3x^2 – 9x – 30

Toán Lớp 8: phân tích đa thức thành nhân tử
9x^2 – 16y^2 – 6x + 1
x^2+5x-24
xy-4a^2xy
3x^2 – 5x – 2
3x^2 – 9x – 30

Comments ( 2 )

  1. #andy
    \[\begin{array}{l}
    a)9{x^2} – 16{y^2} – 6x + 1\\
     = {3^2}.{x^2} – 2.3x.1 + 1 – {4^2}.{y^2}\\
     = {(3x)^2} – 2.3x.1 + 1 – {(4y)^2}\\
     = (3x – 1 – 4y)(3x – 1 + 4y)\\
    b){x^2} + 5x – 24\\
     = {x^2} – 3x + 8x – 24\\
     = ({x^2} – 3x) + (8x – 24)\\
     = x(x – 3) + 8(x – 3)\\
     = (x – 3)(x + 8)\\
    c)9xy – 4{a^2}xy\\
     = xy(9 – 4{a^2})\\
     = xy(3 – 2a)(3 + 2a)\\
    d)3{x^2} – 5x – 2\\
     = 3{x^2} + 6x – x – 2\\
     = (3{x^2} + 6x) – (x + 2)\\
     = 2x(x + 2) – (x + 2)\\
     = (2x – 1)(x + 2)\\
    e)3{x^2} – 9x – 30\\
     = 3{x^2} + 6x – 15x – 30\\
     = (3{x^2} + 6x) – (15x + 30)\\
     = 3x(x + 2) – 15(x + 2)\\
     = (3x – 15)(x + 2)\\
     = 3(x – 5)(x + 2)
    \end{array}\]

  2. Giải đáp + Lời giải và giải thích chi tiết:
    $9x^2 – 16y^2 – 6x + 1$ 
    $3^2.x^2 – 2.3x.1 + 1 – 4^2.y^2$
    $= (3x)^2 – 2.3x.1 + 1 – (4y)^2$
    $= (3x – 1)^2 – (4y)^2$
    $= (3x – 1 – 4y)(3x – 1 + 4y)$
    $x^2+5x-24 $
    $=(x^2-3x)+(8x-24) $
    $= x(x-3)+8(x-3)$
    $=(x-3)(x+8)$
    $9xy-4a^2xy $
    $=xy(9-4a^2)$
    $=xy(3-2a)(3+2a)$
    $3x^2 – 5x – 2$
    $=3x^2 + 6x – x – 2$
    $=(3x^2 + 6x) – (x + 2)$
    $=2x(x + 2) – (x + 2)$
    $=(2x – 1)(x + 2)$
    $3x^2 – 9x – 30$
    $=3x^2 + 6x – 15x – 30$
    $=(3x^2 + 6x) – (15x + 30)$
    $=3x(x + 2) – 15(x + 2)$
    $=(3x – 15)(x + 2)$
    $=3(x – 5)(x + 2)$

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222-9+11+12:2*14+14 = ? ( )