a) 2x– 3 = 3(x – 1) + x + 2 ; b) 2x(x –3) – 5(x –3) = 0 c) 2 2xxx8 x1(x1)(x4)    giả hộ mình vs mình vote cho 5 sao nhố

Question

a) 2x– 3 = 3(x – 1) + x + 2 ; b) 2x(x –3) – 5(x –3) = 0
c)
2
2xxx8
x1(x1)(x4)
 

giả hộ mình vs mình vote cho 5 sao nhố

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Lyla 9 phút 2021-09-16T18:53:04+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-16T18:54:10+00:00

    a, 2x– 3 = 3(x – 1)+x+2

    ⇔2x-3=3x-3+x+2

    ⇔2x-3=4x-1

    ⇔2x-4x=3-1

    ⇔-2x    =2

    ⇔x=-1

    b,2x(x –3) – 5(x –3)=0

    ⇔2x²-6x-5x+15     =0

    ⇔2x²-11x+15         =0

    ⇔(2x²-6x)-(5x-15)   =0

    ⇔2x(x-3)-4(x-3)       =0

    ⇔(x-3)(2x-4)            =0

    ⇔\(\left[ \begin{array}{l}x-3=0\\2x-4=0\end{array} \right.\) 

    ⇔\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\) 

    c,       đkxđ:x$\neq$1;-4

    $\frac{2x}{x+1}$ =$\frac{x²-x+8}{(x-1)(x+4)}$ 

    ⇔$\frac{2x(x+4)}{(x+1)(x-1)}$ =$\frac{x²-x+8}{(x-1)(x+4)}$ 

    ⇔2x(x+4)=x²-x+8

    ⇔2x²+8x=x²-x+8

    ⇔2x²+8x-x²+x-8=0

    ⇔(x²+x)-(8x+8)=0

    ⇔x(x+1)-8(x+1)=0

    ⇔(x+1)(x-8)=0

    ⇔\(\left[ \begin{array}{l}x+1=0\\x-8=0\end{array} \right.\) 

    ⇔\(\left[ \begin{array}{l}x=-1\\x=8\end{array} \right.\) 

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