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

Toán Lớp 9: B1 : giải pt a) 5x-2/3 + x=1 + 5-3x/2 b)(x+2)(3-4x)=x^2 + 4x +4

Toán Lớp 9: B1 : giải pt
a) 5x-2/3 + x=1 + 5-3x/2
b)(x+2)(3-4x)=x^2 + 4x +4

Comments ( 2 )

  1. a/ $\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\\\leftrightarrow 2(5x-2)+6x=6+3(5-3x)\\\leftrightarrow 10x-4+6x=6+15-9x\\\leftrightarrow 16x-4=-9x+21\\\leftrightarrow 25x=25\\\leftrightarrow x=1$
    Vậy pt có tập nghiệm $S=\{1\}$
    b/ $(x+2)(3-4x)=x^2+4x+4\\\leftrightarrow (x+2)(3-4x)=(x+2)^2\\\leftrightarrow (x+2)(3-4x)-(x+2)^2=0\\\leftrightarrow (x+2)[(3-4x)-(x+2)]=0\\\leftrightarrow (x+2)(3-4x-x-2)=0\\\leftrightarrow (x+2)(-5x+1)=0\\\leftrightarrow\left[\begin{array}{1}x+2=0\\-5x+1=0\end{array}\right.\\\leftrightarrow\left[\begin{array}{1}x=-2\\-5x=-1\end{array}\right.\\\leftrightarrow\left[\begin{array}{1}x=-2\\x=\dfrac{1}{5}\end{array}\right.$
    Vậy pt có tập nghiệm $S=\left\{-2;\dfrac{1}{5}\right\}$

  2. Giải đáp + Lời giải và giải thích chi tiết:
    a) (5x-2)/3+x=1+(5-3x)/2
    <=> (2.((5x-2))/6) +(6x)/6 =6/6+(3.((5-3x))/6)
    => 10x-4+6x=6+15-9x
    <=> 25x=25
    <=>x=1
    Vậy x=1 
    b) (x+2)(3-4x)=x^2 + 4x +4
    <=> 3x-4x^2+6-8x=x^2+4x+4
    <=>5x^2 +9x-2<=>5x^2+10x-x-2 =0
    <=> 5x(x+2)-(x+2)=0
    <=> (x+2).(5x-1)=0
    <=> $\left[\begin{matrix} x+2=0\\ 5x-1=0\end{matrix}\right.$
    <=> $\left[\begin{matrix} x=-2\\ x=\dfrac{1}{5}\end{matrix}\right.$
    Vậy x=-2 hoặc x=1/5 
     

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

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