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

Toán Lớp 7: Tìm số hữu tỉ x a) (x-3/2).(2x+1)>0 b)(2-x).(4/5-x)<0 c)(x:7/3).1/7= -3/8 d) (3x+2/5).(x-4/7)=0

Tháng Hai 20, 2023 Toán học 0 Comments 0 views

Toán Lớp 7: Tìm số hữu tỉ x
a) (x-3/2).(2x+1)>0
b)(2-x).(4/5-x)<0 c)(x:7/3).1/7= -3/8 d) (3x+2/5).(x-4/7)=0

Comments ( 1 )

  1. hienthucthu97
    hienthucthu97
    20 Tháng Hai, 2023 at 1:22 chiều
    Reply
    a) (x – 3/2). (2x + 1) > 0
    => x – 3/2 và 2x + 1 cùng dấu
    Ta có 2 trường hợp:
    TH1: \(\left\{\begin{matrix}x – \dfrac{3}{2} > 0\\2x + 1 > 0\end{matrix}\right.\)
    => \(\left\{\begin{matrix}x > \dfrac{3}{2}\\x > \dfrac{-1}{2}\end{matrix}\right.\)
    => x > 3/2
    TH2: \(\left\{\begin{matrix}x – \dfrac{3}{2} < 0\\2x + 1 < 0\end{matrix}\right.\)
    => \(\left\{\begin{matrix}x < \dfrac{3}{2}\\x < \dfrac{-1}{2}\end{matrix}\right.\)
    => x < (-1)/2
    Vậy x > 3/2 hoặc x < (-1)/2
    b) (2 – x). (4/5 – x) < 0
    => 2 – x và 4/5 – x trái dấu
    Ta có 2 trường hợp
    TH1: \(\left\{\begin{matrix}2 – x > 0\\\dfrac{4}{5} – x < 0\end{matrix}\right.\)
    => \(\left\{\begin{matrix}2 > x\\\dfrac{4}{5} < x\end{matrix}\right.\)
    => 4/5 < x < 2
    TH2: \(\left\{\begin{matrix}2 – x < 0\\\dfrac{4}{5} – x > 0\end{matrix}\right.\)
    => \(\left\{\begin{matrix}2 < x\\\dfrac{4}{5} > x\end{matrix}\right.\)
    => 4/5 > x > 2(Vô lí, Loại)
    Vậy 4/5 < x < 2
    c) (x : 7/3). 1/7 = (-3)/8
    => x : 7/3 = (-3)/8 : 1/7
    => x : 7/3 = (-21)/8
    => x = (-21)/8. 7/3
    => x = (-49)/8
    Vậy x = (-49)/8
    d) (3x + 2/5). (x – 4/7) = 0
    => \(\left[ \begin{array}{l}3x + \dfrac{2}{5} = 0\\x- \dfrac{4}{7} = 0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}3x = \dfrac{-2}{5}\\x= \dfrac{4}{7}\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x = \dfrac{-2}{15}\\x = \dfrac{4}{7}\end{array} \right.\) 
    Vậy x in {(-2)/15; 4/7}

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

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