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

Toán Lớp 8: 1 . x^2-3x-7(x^2-9)=0 2. 2x-3+5(2x-3)^2=0 3. 4x^2-4x+1-9(4x^2-1)=0 ( đề bài là tìm x ạ) ( giúp em với )

Tháng Chín 9, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: 1 . x^2-3x-7(x^2-9)=0
2. 2x-3+5(2x-3)^2=0
3. 4x^2-4x+1-9(4x^2-1)=0 ( đề bài là tìm x ạ) ( giúp em với )

Comments ( 2 )

  1. ngoclandiep
    ngoclandiep
    9 Tháng Chín, 2022 at 2:20 sáng
    Reply
    Giải đáp:
    1, x\in{3;-21/6}
    2, x\in{3/2;7/5}
    3, x\in{1/2;-5/8}
    Lời giải và giải thích chi tiết:
     1)
    x^2-3x-7(x^2-9)=0
    <=>x(x-3)-7(x-3)(x+3)=0
    <=>(x-3).[x-7(x+3)]=0
    <=>(x-3).(x-7x-21)=0
    <=>(x-3)(-6x-21)=0
    \(⇔\left[ \begin{array}{l}x-3=0\\-6x-21=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=3\\-6x=21\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=3\\x=-\dfrac{21}{6}\end{array} \right.\)
    Vậy x=3 hoặc x=-21/6
    2)
    2x-3+5(2x-3)^2=0
    ⇔(2x-3)+5(2x-3)^2=0
    ⇔(2x-3).[1+5(2x-3)]=0
    ⇔(2x-3)(1+10x-15)=0
    ⇔(2x-3)(10x-14)=0
    \(⇔\left[ \begin{array}{l}2x-3=0\\10x-14=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}2x=3\\10x=14\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{14}{10}\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{7}{5}\end{array} \right.\)
    Vậy x=3/2 hoặc x=7/5
    3)
    4x^2-4x+1-9(4x^2-1)=0
    ⇔(2x)^2-2.2x.1+1^2-9.[(2x)^2-1]=0
    ⇔(2x-1)^2-9(2x-1)(2x+1)=0
    ⇔(2x-1).[(2x-1)-9(2x+1)]=0
    ⇔(2x-1)(2x-1-18x-9)=0
    ⇔(2x-1)(-16x-10)=0
    \(⇔\left[ \begin{array}{l}2x-1=0\\-16x-10=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{10}{16}\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{5}{8}\end{array} \right.\)
    Vậy x=1/2 hoặc x=-5/8

  2. maianh67
    maianh67
    9 Tháng Chín, 2022 at 2:20 sáng
    Reply
    $\\$
    1,
    x^2 – 3x – 7 (x^2-9)=0
    ⇔ x (x-3)-7(x-3)(x+3)=0
    ⇔ (x-3)(x – 7x – 21)=0
    ⇔ (x-3)(-6x – 21)=0
    ⇔ \(\left[ \begin{array}{l}x-3=0\\-6x-21=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=3\\x=\dfrac{-21}{6}\end{array} \right.\) 
    Vậy x=3,x=(-21)/6
    2,
    2x-3+5(2x-3)^2=0
    <=>(2x-3)+5(2x-3)^2=0
    <=> (2x-3)(1+10x – 15)=0
    <=> (2x-3)(10x – 14)=0
    <=> \(\left[ \begin{array}{l}2x-3=0\\10x-14=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{7}{5}\end{array} \right.\) 
    Vậy x=3/2,x=7/5
    3,
    4x^2 – 4x+1-9(4x^2-1)=0
    <=> ( 2x-1)^2 – 9 (2x-1)(2x+1)=0
    <=> (2x-1)(2x-1 -18x – 9)=0
    <=> (2x-1)(-16x – 10)=0
    <=> \(\left[ \begin{array}{l}2x-1=0\\-16x-10=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{-5}{8}\end{array} \right.\) 
    Vậy x=1/2,x=(-5)/8

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

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