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

Toán Lớp 7: Câu a |x-1|=|2x-3| B |x + 1| + | y-2|=0 c |2x-1| – |x+3|= 0

Toán Lớp 7: Câu a |x-1|=|2x-3|
B |x + 1| + | y-2|=0
c |2x-1| – |x+3|= 0

Comments ( 2 )

  1. $\\$
    a,
    |x-1| = |2x-3|
    -> \(\left[ \begin{array}{l}x-1=2x-3\\x-1=-2x+3\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}x-2x=1-3\\x+2x=1+3\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}-x=-2\\3x=4\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}x=2\\x=4:3\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}x=2\\x=\dfrac{4}{3}\end{array} \right.\) 
    Vậy x=2 hoặc x=4/3
    $\\$
    b,
    |x+1| + |y-2|=0
    Với mọi x,y có : $\begin{cases} |x+1|≥0\\|y-2|≥0 \end{cases}$
    -> |x+1| + |y-2| ≥0∀x,y
    Dấu “=” xảy ra khi :
    ↔ $\begin{cases} |x+1|=0\\|y-2|=0 \end{cases}$
    ↔ $\begin{cases} x+1=0\\y-2=0 \end{cases}$
    ↔ $\begin{cases} x=0-1\\y=0+2\end{cases}$
    ↔ $\begin{cases} x=-1\\y=2\end{cases}$
    Vậy (x;y)=(-1;2)
    $\\$
    c,
    |2x-1| -|x+3|=0
    -> |2x-1|=0+|x+3|
    -> |2x-1|=|x+3|
    -> \(\left[ \begin{array}{l}2x-1=x+3\\2x-1=-x-3\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}2x-x=1+3\\2x+x=1-3\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}x=4\\3x=-2\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}x=4\\x=-2:3\end{array} \right.\) $\\$ -> \(\left[ \begin{array}{l}x=4\\x=\dfrac{-2}{3}\end{array} \right.\) 
    Vậy x=4 hoặc x=(-2)/3
     

  2. Giải đáp+Lời giải và giải thích chi tiết:
    a)|x-1|=|2x-3|
    <=>[(x-1=2x-3),(x-1=3-2x):}
    <=>[(x=2),(3x=4):}
    <=>[(x=2),(x=4/3):}
    Vậy x=2 hoặc x=4/3.
    b)|x+1|+|y-2|=0
    Vì |x+1|>=0
    \qquad |y-2|>=0
    =>|x+1|+|y-2|>=0
    Mà đề bài cho |x+1|+|y-2|=0
    <=>{(x+1=0),(y-2=0):}
    <=>{(x=-1),(y=2):}
    c)|2x-1|-|x+3|=0
    <=>|2x-1|=|x+3|
    <=>[(2x-1=x+3),(2x-1=-x-3):}
    <=>[(x=4),(3x=-2):}
    <=>[(x=4),(x=-2/3):}
    Vậy x=4 hoặc x=-2/3.

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

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