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

Toán Lớp 6: b) -6 – 3x + 60 = (-2) mũ 2. (-3) c) (x – 2) mũ 2 = 4. (-5) mũ 2 d) (x mũ 2 + 9). (x mũ 3 + 8) = 0 Giup Mik thật nhanh nhé!!!! Mik sẽ

Tháng Sáu 20, 2022 Toán học 2 Comments 0 views

Toán Lớp 6: b) -6 – 3x + 60 = (-2) mũ 2. (-3)
c) (x – 2) mũ 2 = 4. (-5) mũ 2
d) (x mũ 2 + 9). (x mũ 3 + 8) = 0
Giup Mik thật nhanh nhé!!!! Mik sẽ cho câu trl hay nhất

Comments ( 2 )

  1. khanhlanchau
    khanhlanchau
    20 Tháng Sáu, 2022 at 10:11 sáng
    Reply
    Answer
    b, -6 – 3x + 60 = (-2)^{2} . (-3)
    => -6 + 60 – 3x = 4 . (-3)
    => 54 – 3x = -12
    => 3x = 54 + 12
    => 3x = 66
    => x = 66 : 3
    => x = 22
    __________________________
    c, (x – 2)^{2} = 4 . (-5)^{2}
    => (x – 2)^{2} = 4 . 25
    => (x – 2)^{2} = 100
    => (x – 2)^{2} = (+-10)^{2}
    => x – 2 = +-10
    => $\left[\begin{matrix} x – 2 = 10\\ x – 2 = -10\end{matrix}\right.$
    => $\left[\begin{matrix} x = 10 + 2\\ x = -10 + 2\end{matrix}\right.$
    => $\left[\begin{matrix} x = 12\\ x = -8\end{matrix}\right.$
    $\text{Vậy}$ x \in {12 ; -8}
    ____________________________
    d, (x^{2} + 9) . (x^{3} + 8) = 0
    => $\left[\begin{matrix} x^2 + 9 = 0\\ x^3 + 8 = 0\end{matrix}\right.$
    => $\left[\begin{matrix} x^2 = 0 – 9\\ x^3 = 0 – 8\end{matrix}\right.$
    => $\left[\begin{matrix} x^2 = – 9 \ \text{(loại)}\\ x^3 = – 8\end{matrix}\right.$
    => $\left[\begin{matrix} x^2 = – 9 \ \text{(loại)}\\ x^3 = (-2)^{3}\end{matrix}\right.$
    => $\left[\begin{matrix} x^2 = – 9 \ \text{(loại)}\\ x = -2 \ \text{(chọn)}\end{matrix}\right.$
    $\text{Vậy}$ x = -2

  2. chauhoangmy98
    chauhoangmy98
    20 Tháng Sáu, 2022 at 10:11 sáng
    Reply
    b) -6 – 3x + 60 = (-2)^2. (-3)
    -6-3x+60=4.(-3)
    -6-3x+60=-12
    3x+60=-6-(-12)
    3x+60=6
    3x=6-60
    3x=-54
    x=(-54):3
    x=-18
    c) (x – 2)^2 = 4. (-5)^2
    (x-2)^2=4.25
    (x-2)^2=100
    (x-2)^2=+-10^2
    =>x-2=10
    =>x=10+2
    =>x=12
    =>\(\left[ \begin{array}{l}x-2=10\\x-2=-10\end{array} \right.\) 
    =>\(\left[ \begin{array}{l}x=12\\x=-8\end{array} \right.\) 
    d) (x^2 + 9). (x^3 + 8) = 0
    =>x^2+9=x^3+8
    =>\(\left[ \begin{array}{l}x^2+9=0\\x^3+8=0\end{array} \right.\) 
    =>\(\left[ \begin{array}{l}x^2=-9\\x^3=-8\end{array} \right.\) 
    =>\(\left[ \begin{array}{l}x^2=\pm3^2\\x^3=\pm2^3\end{array} \right.\) 
    =>\(\left[ \begin{array}{l}x=\pm3\\x=\pm2\end{array} \right.\) 
     

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

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