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

Toán Lớp 8: Tìm x biết : a, 7x( x – 5) -x + 5 = 0 b, ( x – 3)^2 – 4( x + 2 ) ^2 = 0

Toán Lớp 8: Tìm x biết :
a, 7x( x – 5) -x + 5 = 0
b, ( x – 3)^2 – 4( x + 2 ) ^2 = 0

Comments ( 2 )

  1. Giải đáp:
     a, S = {1/7 ; 5}
    b, S = {-7 ; -1/3}
    Lời giải và giải thích chi tiết:
     a, 7x(x – 5) – x + 5 = 0
    <=> 7x(x – 5) – (x – 5) = 0
    <=> (7x – 1)(x – 5) = 0
    ⇔\(\left[ \begin{array}{l}7x – 1 = 0\\x – 5 = 0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x = \frac{1}{7}\\x = 5\end{array} \right.\) 
    Vậy S = {1/7 ; 5}
    b, (x – 3)^2 – 4(x + 2)^2 = 0
    <=> (x – 3)^2 – ( 2(x + 2))^2 = 0
    <=> (x – 3 – 2x – 4)(x – 3 + 2x + 4) = 0
    <=> (-x – 7)(3x + 1) = 0
    ⇔\(\left[ \begin{array}{l}-x-7=0\\3x+1=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=-7\\x=-\frac{1}{3}\end{array} \right.\) 
    Vậy S = {-7 ; -1/3}

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a ) 7 x ( x − 5 ) − x + 5 = 0
    ⇔ 7 x ( x − 5 ) − ( x − 5 ) = 0
    ⇔( x − 5 ) ( 7 x − 1 ) = 0
    ⇔ $\left[\begin{matrix} x − 5 = 0\\7 x − 1 = 0\end{matrix}\right.$
    ⇔ $\left[\begin{matrix} x = 5\\x = \dfrac{1}{7}\end{matrix}\right.$
    Vậy S = {1/7;5}
    b) ( x − 3 )^ 2 − 4 ( x + 2 ) ^2 = 0
    ⇔ ( x − 3 )^ 2 − [ 2 ( x + 2 ) ]^2 = 0
    ⇔ ( x − 3 )^ 2 − ( 2 x + 4 ) ^2 = 0
    ⇔ [( x − 3 − (2 x + 4 )][ ( x − 3 + (2 x + 4 )] = 0
    ⇔ ( x − 3 − 2 x – 4 )[ ( x − 3 + 2 x + 4 ) = 0
    ⇔ ( − x − 7 ) ( 3 x + 1 ) = 0
    ⇔ (  x + 7 ) ( 3 x + 1 ) = 0
    ⇔ $\left[\begin{matrix} x + 7 = 0\\ 3 x + 1 = 0\end{matrix}\right.$
    ⇔ $\left[\begin{matrix} x = -7\\x = \dfrac{-1}{3}\end{matrix}\right.$
    Vậy S = {-7;1/3}
    ___________________________
    Áp dụng:
    HĐT: 3,A^2-B^2=(A-B)(A+B)
     

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