Bài 1: Tìm x, bt a) 5\12.|x|=10\3 b) 7\3 – 2.|x|= 1,5 c) |4 – x| \2và3\5=-3\5 d) |x-1\2| + 5 = 14,5

Question

Bài 1: Tìm x, bt
a) 5\12.|x|=10\3
b) 7\3 – 2.|x|= 1,5
c) |4 – x| \2và3\5=-3\5
d) |x-1\2| + 5 = 14,5

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Genesis 1 tuần 2021-12-02T19:41:22+00:00 2 Answers 3 views 0

Answers ( )

    0
    2021-12-02T19:42:34+00:00

    Giải thích các bước giải:

    a) $\frac{5}{12}$ . | x | = $\frac{10}{3}$ 

    | x | = $\frac{10}{3}$ . $\frac{12}{5}$ 

    | x | = 8 

    →\(\left[ \begin{array}{l}x=8\\x=-8\end{array} \right.\)

    Vậy x ∈ { ± 8 }

    b) $\frac{7}{3}$ – 2 . l x l = 1,5

    → $\frac{7}{3}$ – 2 . l x l = $\frac{3}{2}$ 

    → 2 . | x | = $\frac{7}{3}$ – $\frac{3}{2}$ 

    → 2 . | x | = $\frac{5}{6}$ 

    | x | = $\frac{5}{6}$ . $\frac{1}{2}$ 

    | x | = $\frac{5}{12}$ 

    → \(\left[ \begin{array}{l}x=5/12\\x=-5/12\end{array} \right.\) 

     Vậy x ∈ { ± $\frac{5}{12}$ }

    c) $\frac{|4 – x|}{2}$ + $\frac{3}{5}$ = $\frac{-3}{5}$

    |4 – x| . $\frac{1}{2}$ = $\frac{-3}{5}$ – $\frac{3}{5}$

    →  |4 – x| = $\frac{-6}{5}$ . 2

    →  |4 – x| = $\frac{-12}{5}$ ( loại )

    d) | $\frac{x}{1/2}$ | + 5 = 14,5

    → | $\frac{x}{1/2}$ | = $\frac{19}{2}$ 

    →\(\left[ \begin{array}{l}x- 1/2 = 19 /2\\x-1/2=-19/2\end{array} \right.\) 

    → \(\left[ \begin{array}{l}x=10\\x=-9\end{array} \right.\) 

    Vậy x ∈ { 10 ; -9 }

    0
    2021-12-02T19:43:18+00:00

    Đáp án:

     d) \(\left[ \begin{array}{l}
    x = 10\\
    x =  – 9
    \end{array} \right.\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    a)\dfrac{5}{{12}}.\left| x \right| = \dfrac{{10}}{3}\\
     \to \left| x \right| = 8\\
     \to \left[ \begin{array}{l}
    x = 8\\
    x =  – 8
    \end{array} \right.\\
    b)\dfrac{7}{3} – 2\left| x \right| = \dfrac{3}{2}\\
     \to 2\left| x \right| = \dfrac{5}{6}\\
     \to \left| x \right| = \dfrac{5}{{12}}\\
     \to \left[ \begin{array}{l}
    x = \dfrac{5}{{12}}\\
    x =  – \dfrac{5}{{12}}
    \end{array} \right.\\
    d)\left| {x – \dfrac{1}{2}} \right| + 5 = \dfrac{{29}}{2}\\
     \to \left| {x – \dfrac{1}{2}} \right| = \dfrac{{19}}{2}\\
     \to \left[ \begin{array}{l}
    x – \dfrac{1}{2} = \dfrac{{19}}{2}\\
    x – \dfrac{1}{2} =  – \dfrac{{19}}{2}
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    x = 10\\
    x =  – 9
    \end{array} \right.
    \end{array}\)

    ( câu c đề không rõ ràng b nhé )

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