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

Toán Lớp 8: Giải phương trình sau: a) 2|5x| – 3=4 b) 7/2 -3|5-2x|=9/2 c)|3x-1|+4=6-|3x-1|

Toán Lớp 8: Giải phương trình sau:
a) 2|5x| – 3=4
b) 7/2 -3|5-2x|=9/2
c)|3x-1|+4=6-|3x-1|

Comments ( 2 )

  1. $@Mốc$
    a)  $2$.|$5.x$| – $3$ = $4$
    ⇔ $2$.|$5.x$| = $7$
    ⇔        |$5.x$| = $\frac{7}{2}$ 
    ⇔ \(\left[ \begin{array}{l}5x=\frac{7}{2}\\5x=\frac{-7}{2}\end{array} \right.\)
    ⇔ \(\left[ \begin{array}{l}x=\frac{7}{10}\\x=\frac{-7}{10}\end{array} \right.\)
    Vậy phương trình có nghiệm $x$ = $\frac{7}{10}$ và $x$ = $\frac{-7}{10}$
    b)  $\frac{7}{2}$ – $3$.|$5 – 2x$| = $\frac{9}{2}$ 
    ⇔                   $3$.|$5 – 2x$| =    $-1$
    ⇔                         |$5 – 2x$| =$\frac{-1}{3}$ (vô lý)
    Vậy phương trình vô nghiệm.
    c)  |$3x – 1$| + $4$ = $6$ – |$3x – 1$|
    ⇔ $2$.$|3x – 1|$ = $2$
    ⇔        $|3x – 1|$ = $1$
    ⇔  \(\left[ \begin{array}{l}3x – 1 = 1\\3x – 1=-1\end{array} \right.\)
    ⇔  \(\left[ \begin{array}{l}3x = 2\\3x = 0\end{array} \right.\)
    ⇔  \(\left[ \begin{array}{l}x =\frac{2}{3}\\x = 0\end{array} \right.\)
    Vậy phương trình có nghiệm $x$ = $\frac{2}{3}$ và $x$ = $0$
    $#chucbanhoctotnhe;333$  

  2. a):
    2|5x|-3=4
    ⇔ 2|5x|−3=4
    ⇔ 2|5x|=7
    ⇒ |5x| = 7/2
    ⇒ \(\left[ \begin{array}{l}5x={ \dfrac{7}{2}}\\5x=-{ \dfrac{7}{2}}\end{array} \right.\) 
    \(\left[ \begin{array}{l}x={ \dfrac{7}{10}}\\x=-{ \dfrac{7}{10}}\end{array} \right.\) 
    b):
    7/2-3|5-2x|=9/2
    ⇔ 7/2 – 3 × |5 – 2x| = 9/2
    ⇔ 7/2 – 3 × |-2x + 5| = 9/2
    ⇔ (7 – 2(3 × |-2x + 5|))/2 = 9/2
    ⇔ (7 – 2 × 3 × |-2x + 5|)/2 = 9/2
    ⇔ (7 – 6 × |-2x + 5|)/2 = 9/2
    ⇔ (-6 × |-2x + 5| + 7)/2 = 9/2
    ⇔ -6 × |-2x + 5| + 7 = 9
    ⇔ -6 × |-2x + 5| = 2
    ⇔ |-2x + 5| = -2/(2 × 3)
    ⇔ |-2x + 5| = -1/3
    x ∈ \emptyset
    c):
    |3x – 1| + 4 = 6 – |3x – 1|
    ⇔ |3x-1|+4=-|3x-1|+6
    ⇔ 2 × |3x – 1| = 2
    x ∈ {2/3, 0}
     

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