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

Toán Lớp 7: Tìm x biết: a, $x^{2}$ +x=0 b, $(2x+1)^{2}$ =25 c, $(2x-3)^{2}$ =36 d,$(2x-1)^{3}$ =-8 e, $(x-1)^{x-2}$ = $(x-1)^{x-4}$

Tháng Một 4, 2023 Toán học 2 Comments 0 views

Toán Lớp 7: Tìm x biết:
a, $x^{2}$ +x=0
b, $(2x+1)^{2}$ =25
c, $(2x-3)^{2}$ =36
d,$(2x-1)^{3}$ =-8
e, $(x-1)^{x-2}$ = $(x-1)^{x-4}$

Comments ( 2 )

  1. huenthanh97
    huenthanh97
    4 Tháng Một, 2023 at 8:45 sáng
    Reply
    a) x^{2}+x=0
    ->x(x+1)=0
    ->\(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\)
    ->\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\) 
    b) (2x+1)^{2}=25
    ->(2x+1)^{2}=5^{2}
    ->2x+1=5
    ->2x=4
    ->x=2
    c) (2x-3)^{2}=36
    ->(2x-3)^{2}=6^{2}
    ->2x-3=6
    ->2x=9
    ->x=9/2
    d) (2x-1)^{3}=-8
    ->(2x-1)^{3}=-2^{3}
    ->2x-1=-2
    ->2x=-1
    ->x=-1/2
    e) (x-1)^{x-2}=(x-1)^{x-4}
    ->(x-1)^{x-2}:(x-1)^{x-4}=1
    ->(x-1)^{x-2-x+4}=1
    ->(x-1)^{2}=1
    ->(x-1)^{2}=1^{2}
    TH1:
    ->x-1=1
    ->x=2 
    TH2:
    ->x-1=-1
    ->x=0

  2. daithanh
    daithanh
    4 Tháng Một, 2023 at 8:45 sáng
    Reply
    a)  x^2 +x =0
    => x. (x+1)=0
    =>$\left[\begin{matrix} x=0\\ x+1=0\end{matrix}\right.$
    => $\left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.$
    b) (2x+1)^2 =25
    (2x+1)^2=(±5)^2
    =>$\left[\begin{matrix} 2x+1=5\\ 2x+1=-5\end{matrix}\right.$
    =>$\left[\begin{matrix} 2x=4\\ 2x=-6\end{matrix}\right.$
    =>$\left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.$
    c) (2x-3)^2=36
    (2x-3)^2=(±6)^2
    => $\left[\begin{matrix} 2x-3=6\\ 2x-3=-6\end{matrix}\right.$
    => $\left[\begin{matrix} 2x=9\\ 2x=-3\end{matrix}\right.$
    => $\left[\begin{matrix} x=9/2\\ x=-3/2\end{matrix}\right.$
    d)  (2x-1)^3=-8
    (2x-1)^3=-2^3
    => 2x-1 =-2
    => 2x =-1
    => x =-1/2
    e)  (x-1)^(x-2)=(x-1)^(x-4)
    (x-1)^(x-2) – (x-1)^{(x-4)}=0
    (x-1)^(x-4).[(x-1)^2 -1]=0
    => $\left[\begin{matrix} (x-1)^{(x-4)} =0\\ (x-1)^2 -1=0\end{matrix}\right.$
    => $\left[\begin{matrix} x-1=0\\ (x-1)^2=1\end{matrix}\right.$
    => $\left[\begin{matrix} x=1\\ (x-1)^2=(±1)^2\end{matrix}\right.$
    => $\left[\begin{matrix} x=1\\ x-1=1 ;x-1 =-1\end{matrix}\right.$
    => $\left[\begin{matrix} x=1\\ x=2;x=0\end{matrix}\right.$

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

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