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

Toán Lớp 8: Tìm x a. x(x-3)-x+3=0 b. (6-x)(2x+9)=45 c. 4x^2 -12x+9=0

Toán Lớp 8: Tìm x
a. x(x-3)-x+3=0
b. (6-x)(2x+9)=45
c. 4x^2 -12x+9=0

Comments ( 2 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
     a) x(x-3)-x+3=0
    <=> x(x-3)-(x-3)=0
    <=> (x-1)(x-3)=0
    <=> [(x-1=0),(x-3=0):}
    <=> [(x=1),(x=3):}
    Vậy x \in {1;3}
    b) (6-x)(2x+9)=45
    <=> 6(2x+9)-x(2x+9)=45
    <=> 12x+54-2x^2-9x=45
    <=> -2x^2+3x+54=45
    <=> -2x^2+3x+54-45=0
    <=> -2x^2+3x+9=0
    <=> -(2x^2-3x-9)=0
    <=> 2x^2-3x-9=0
    <=> 2(x^2-3/2x-9/2)=0
    <=> x^2-3/2x-9/2=0
    <=> (x^2-3/2x+9/16)-81/16=0
    <=> (x-3/4)^2-(9/4)^2=0
    <=> (x-3/4-9/4)(x-3/4+9/4)=0
    <=> (x-3)(x+3/2)=0
    <=> [(x-3=0),(x+3/2=0):}
    <=> [(x=3),(x=-3/2):}
    Vậy x \in {3;-3/2}
    c) 4x^2-12x+9=0
    <=> (2x)^2-2.2x.3+3^2=0
    <=> (2x-3)^2=0
    <=> 2x-3=0
    <=> 2x=3
    <=> x=3/2
    Vậy x=3/2

  2. Giải đáp + Lời giải và giải thích chi tiết :
    a) x ( x – 3 ) – x + 3 = 0
    ⇔ x ( x – 3 ) – ( x – 3 ) = 0
    ⇔ ( x – 3 ) ( x – 1 ) = 0
    ⇔ \(\left[ \begin{array}{l}x-3=0\\x-1=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\) 
    Vậy x = 3 ; x =  1
    b) ( 6 – x ) ( 2x + 9 ) = 45
    ⇔ 12x + 54 – 2x^2 – 9x = 45
    ⇔ 3x + 54 – 2x^2 = 45
    ⇔ 3x + 54 – 2x^2 – 45 = 0
    ⇔ 2x^2 – 3x – 9 = 0
    ⇔ 2x^2 + 3x – 6x – 9 = 0
    ⇔ x ( 2x + 3 ) – 3 ( 2x + 3 ) = 0
    ⇔ ( 2x + 3 ) ( x – 3 ) = 0
    ⇔ \(\left[ \begin{array}{l}2x+3=0\\x-3=0\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}2x=-3\\x=3\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l}x=-\frac{3}{2}\\x=3\end{array} \right.\) 
    Vậy x = – 3/2 ; x = 3
    c) 4x^2 – 12x + 9 = 0
    ⇔ ( 2x – 3 )^2 = 0
    ⇔ 2x – 3 = 0
    ⇔ 2x = 3
    ⇔ x = 3/2
    Vậy x = 3/2
    $# Mon $
    $# Xin Hay Nhất $

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

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