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

Toán Lớp 8: 1)2x(x-5)+20-4x=0 2)(2xy-2)(2xy+3)-(1-2xy)^2 tại x=1/2 , y=-1 3)6x^3-10xy

Toán Lớp 8: 1)2x(x-5)+20-4x=0 2)(2xy-2)(2xy+3)-(1-2xy)^2 tại x=1/2 , y=-1 3)6x^3-10xy

Comments ( 2 )

  1. Giải đáp:
    1) x = 5 hoặc x = 2
    2) – 10
    3) 2x(3x^2 – 5y)
    Lời giải và giải thích chi tiết:
    1)
    2x(x – 5) + 20 – 4x = 0
    2x(x – 5) + 4(5 – x) = 0
    2x(x – 5) – 4(x – 5) = 0
    (x – 5)(2x – 4) = 0
    2(x – 5)(x – 2) = 0
    => x – 5 = 0 hoặc x – 2 = 0
    => x = 5 hoặc x = 2
    Vậy x = 5 hoặc x = 2
    * Nhóm hạng tử, đặt nhân tử chung.
    2)
    (2xy – 2)(2xy + 3) – (1 – 2xy)^2 tại x = 1/2, y = – 1
    (4x^2 y^2 + 6xy – 4xy – 6) – [1^2 – 2.1.2xy + (2xy)^2]
    = (4x^2 y^2 + 2xy – 6) – (1 – 4xy + 4x^2 y^2)
    = 4x^2 y^2 + 2xy – 6 – 1 + 4xy – 4x^2 y^2
    = (4x^2 y^2 – 4x^2 y^2) + (2xy + 4xy) – (6 + 1)
    = 6xy – 7
    Thay x = 1/2, y = -1 ta có:
    6 . 1/2 . ( – 1) – 7
    = 3 . ( – 1) – 7
    = – 3 – 7 = – 10
    Vậy (2xy – 2)(2xy + 3) – (1 – 2xy)^2 = – 10 khi x = 1/2, y = – 1
    * Dùng HĐT 2: (a – b)^2 = a^2 – 2ab + b^2
    3)
    6x^3 – 10xy
    = 2x.3x^2 – 2x.5y
    = 2x(3x^2 – 5y)
    * Đặt nhân tử chung.
    #SunHee

  2. Giải đáp:
    $1)$
    $a) 2x.(x-5)+20-4x=0$
    $⇔ 2x.(x-5)+4.(5-x)=0$
    $⇔ 2x.(x-5)-4.(x-5)=0$
    $⇔ (2x-4).(x-5)=0$
    $⇔$ \(\left[ \begin{array}{l}2x-4=0\\x-5=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}2x=4\\x=5\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\) 
    $Vậy$ $x=2$ $hoặc$ $x=5$
    $2) (2xy-2).(2xy+3)-(1-2xy)²$
    $= 4x²y²+6xy-4xy-6-[1²-2.1.2xy+(2xy)²]$
    $= 4x²y²+2xy-6-1+4xy-4x²y²$
    $= (4x²y²-4x²y²)+(2xy+4xy)-(6+1)$
    $= 6xy-7$
    $+)$ $Thay$ $x=\frac{1}{2};y=-1$ $vào$ $bt$ $trên$ $ta$ $đc:$
    $→ 6.\frac{1}{2}.(-1)-7=-10$
    $Vậy$ $giá$ $trị$ $của$ $bt$ $trên$ $khi$ $x=\frac{1}{2};y=-1$ $là$ $(-10)$
    $3) 6x³-10xy$
    $= 2x.(3x²-5y)$
     

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

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