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

Toán Lớp 8: tìm x biết a,3x.(x-2)+x-2=0 b,2.(x+2)-x^2-3x=0 c,4x^2-25-(2x-5).(2x+7)=0 d,x^3+27+(x+3).(x-9)=0 e,5x.(3x+4)-6x-8=0 g,(x-2).(x^2+2x+7)+

Home/ Toán Lớp 8: tìm x biết a,3x.(x-2)+x-2=0 b,2.(x+2)-x^2-3x=0 c,4x^2-25-(2x-5).(2x+7)=0 d,x^3+27+(x+3).(x-9)=0 e,5x.(3x+4)-6x-8=0 g,(x-2).(x^2+2x+7)+

Toán Lớp 8: tìm x biết a,3x.(x-2)+x-2=0 b,2.(x+2)-x^2-3x=0 c,4x^2-25-(2x-5).(2x+7)=0 d,x^3+27+(x+3).(x-9)=0 e,5x.(3x+4)-6x-8=0 g,(x-2).(x^2+2x+7)+

Tháng Mười Hai 9, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: tìm x biết
a,3x.(x-2)+x-2=0
b,2.(x+2)-x^2-3x=0
c,4x^2-25-(2x-5).(2x+7)=0
d,x^3+27+(x+3).(x-9)=0
e,5x.(3x+4)-6x-8=0
g,(x-2).(x^2+2x+7)+2.(x^2-4)-5.(x-2)=0
giúp mình vs mình cần gấp

Comments ( 2 )

  1. kimnguenoanh
    kimnguenoanh
    9 Tháng Mười Hai, 2022 at 2:35 sáng
    Reply
    Giải đáp:
     
    Lời giải và giải thích chi tiết:
     a) <=>(x-2)(3x+1)=0
    <=>x-2=0;3x+1=0
    <=>x=2;x=-1/3
    b) 3x->2x
    2(x+2)-x^2-2x=0
    <=>2(x+2)-x(x+2)=0
    <=>(x+2)(2-x)=0
    <=>x=+-2
    c) $4x^2-25-\left(2x-5\right)\cdot \left(2x+7\right)=0$
    $⇔4x^2-\left(2x-5\right)\left(2x+7\right)=25$
    $⇔4x^2-4x^2-4x+35=25$
    $⇔-4x+35=25$
    $⇔-4x=-10$
    ⇔x=\frac{5}{2}
    d) $⇔x^3+x^2-6x=0$
    $⇔x\left(x-2\right)\left(x+3\right)=0$
    $⇒x=0,\:x=2,\:x=-3$
    e) $5x\left(3x+4\right)-6x-8=0$
    $⇔15x^2+14x-8=0$
    $⇔5x(4x+3)-2(4x+3)=0$
    $⇔(4x+3)(5x-2)=0$
    $⇔x=\cfrac{2}{5},\:x=-\cfrac{4}{3}$
    f) $⇔x^3+3x-14+2x^2-8-5\left(x-2\right)=0$
    $⇔x^3+3x-14+2x^2-8-5\left(x-2\right)=0$
    $⇔x^3+3x-14+2x^2-8-5x+10=0$
    $⇔x^3+2x^2-2x-12=0$
    $⇔\left(x-2\right)\left(x^2+4x+6\right)=0$
    $⇔x=2$ do x^2+4x+6>0

  2. bichhhathu
    bichhhathu
    9 Tháng Mười Hai, 2022 at 2:35 sáng
    Reply
    $a) 3x(x-2)+(x-2)=0$
    $⇔(3x+1)(x-2)=0$
    $⇔\left[ \begin{array}{1}3x=-1\\x=2\end{array} \right.$
    $⇔\left[ \begin{array}{1}x=\dfrac{-1}{3}\\x=2\end{array} \right.$
    Vậy $x ∈ \left \{ \dfrac{-1}{3} ; 2 \right \}$
    Câu $b)$ có vẻ sai đề
    $c) 4x^2-25-(2x-5)(2x+7)=0$
    $⇔(2x-5)(2x+5)-(2x-5)(2x+7)=0$
    $⇔(2x-5)(2x+5-2x-7)=0$
    $⇔(2x-5)(-2)=0$
    $⇔2x=5$
    $⇔x=\dfrac{5}{2}$
    Vậy $x=\dfrac{5}{2}$
    $d) x^3+27+(x+3)(x-9)=0$
    $⇔(x+3)(x^2-3x+9)+(x+3)(x-9)=0$
    $⇔(x+3)(x^2-3x+9+x-9)=0$
    $⇔(x+3)(x^2-2x)=0$
    $⇔(x+3)x(x-2)=0$
    $⇔\left[ \begin{array}{1}x=-3\\x=0\\x=2\end{array} \right.$
    Vậy $x ∈ \{-3;0;2\}$
    $e) 5x(3x+4)-6x-8=0$
    $⇔5x(3x+4)-2(3x+4)=0$
    $⇔(5x-2)(3x+4)=0$
    $⇔\left[ \begin{array}{1}5x=2\\3x=-4\end{array} \right.$
    $⇔\left[ \begin{array}{1}x=\dfrac{2}{5}\\x=\dfrac{-4}{3}\end{array} \right.$
    Vậy $x ∈ \left \{ \dfrac{2}{5} ; \dfrac{-4}{3} \right \}$
    $g) (x-2)(x^2+2x+7)+2(x^2-4)-5(x-2)=0$
    $⇔(x-2)(x^2+2x+7-5)+2(x+2)(x-2)=0$
    $⇔(x-2)(x^2+2x+2+2x+4)=0$
    $⇔(x-2)(x^2+4x+6)=0$
    $⇔x-2=0$ ( vì $x^2+4x+6 >0 ∀x$ )
    $⇔x=2$
    Vậy $x=2$
     

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

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