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

Toán Lớp 8: a/ 3x( 2x +7) – ( x+ 1)( 6x -5) = 10 b/ (x – 4)²– 36 =0 c/3(2x – 1) – 5( x – 3) =4 d/ (2x – 1)²–

Tháng Sáu 19, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: a/ 3x( 2x +7) – ( x+ 1)( 6x -5) = 10
b/ (x – 4)²– 36 =0
c/3(2x – 1) – 5( x – 3) =4
d/ (2x – 1)²– (x+3)2²= 0
Đề bài:Tìm x

Comments ( 2 )

  1. dantam45
    dantam45
    19 Tháng Sáu, 2022 at 10:46 sáng
    Reply
    Giải đáp+Lời giải và giải thích chi tiết:
    a) 3x(2x+7)-(x+1)(6x-5)=10
    ⇔6x^2+21x-(6x^2-5x+6x-5)=10
    ⇔6x^2+21x-6x^2+5x-6x+5=10
    ⇔(6x^2-6x^2)+(21x+5x-6x)+5=10
    ⇔20x+5=10
    ⇔20x=5
    ⇔x=1/4
       Vậy S={1/4}
    b) (x-4)^2-36=0
    ⇔(x-4)^2-6^2=0
    ⇔(x-4-6)(x-4+6)=0
    ⇔(x-10)(x+2)=0
    $⇔\left[\begin{matrix} x-10=0\\ x+2=0\end{matrix}\right.$
    $⇔\left[\begin{matrix} x=10\\ x-2\end{matrix}\right.$
        Vậy S={10; -2}
    c) 3(2x-1)-5(x-3)=4
    ⇔6x-3-5x+15=4
    ⇔(6x-5x)-(3-15)=4
    ⇔x+12=4
    ⇔x=-8
       Vậy S={-8}
    d) (2x-1)^2-(x+3)^2=0    (sửa đề)
    ⇔(2x-1-x-3)(2x-1+x+3)=0
    ⇔(x-4)(3x+2)=0
    $⇔\left[\begin{matrix} x-4=0\\ 3x+2=0\end{matrix}\right.$
    $⇔\left[\begin{matrix} x=4\\ x=\dfrac{-2}{3}\end{matrix}\right.$
        Vậy S={4; -2/3}
         Chúc bn hc tốt????
             $\text{#LinhSad}$ 

  2. maianhnhiquynh
    maianhnhiquynh
    19 Tháng Sáu, 2022 at 10:46 sáng
    Reply
    $\#Lynn2007$
    a)3x(2x+7)-(x+1)(6x-5)=10
    <=>6x^2+21x-(6x^2-5x+6x-5)=10
    <=>6x^2+21x-6x^2+5x-6x+5=10
    <=>32x+5=10
    <=>20x=5
    <=>x=5/20
    <=>x=1/4
    \text{Vậy S = } {1/4}
    b)(x-4)^2-36=0
    <=>(x-4)^2-6^2=0
    <=>(x-4-6)(x-4+6)=0
    <=>(x-10)(x+2)=0
    <=> \(\left[ \begin{array}{l}x-10=0\\x+2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=10\\x=-2\end{array} \right.\) 
    \text{Vậy S =} {10;-2}
    c)3(2x-1)-5(x-3)=4
    <=>6x-3-5x+15=4
    <=>x+12=4
    <=>x=-8
    \text{Vậy S =} {-8}
    d)(2x-1)^2-(x+3)^2=0
    <=>(2x)^2-2.2x.1+1^2-(x^2+2.x.3+3^2)=0
    <=>4x^2-4x+1-x^2-6x-9=0
    <=>3x^2-10x-8=0
    <=>3x^2+2x-12x-8=0
    <=>x(3x+2)-4(3x+2)=0
    <=>(x-4)(3x+2)=0
    <=> \(\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=4\\x=\dfrac{-2}{3}\end{array} \right.\) 
    \text{Vậy S =} {4;-2/3}
    \text{—————————————-}
    \text{Áp dụng HĐT:}
    1)a^2-b^2=(a-b)(a+b)
    2)(a+b)^2=a^2+2ab+b^2
    3)(a-b)^2=a^2-2ab+b^2

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

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