Tìm x a) 5(x +3) – 2x(3+x)=0 b) 4x (x-2004) – x + 2004 =0 c) (x + 1)^2 = x+1

Question

Tìm x
a) 5(x +3) – 2x(3+x)=0
b) 4x (x-2004) – x + 2004 =0
c) (x + 1)^2 = x+1

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Eloise 2 tháng 2021-08-09T16:06:30+00:00 2 Answers 4 views 0

Answers ( )

    0
    2021-08-09T16:08:04+00:00

    a) `5(x +3) – 2x(3+x)=0`

    ⇔ `(5-2x)(x+3)=0`

    ⇔ \(\left[ \begin{array}{l}5-2x=0\\x+3=0\end{array} \right.\)

    ⇔ \(\left[ \begin{array}{l}x=5/2\\x=-3\end{array} \right.\)

    b) `4x(x-2004)-x+2004=0`

    ⇔ `4x(x-2004)-(x-2004)=0`

    ⇔ `(4x-1)(x-2004)=0`

    ⇔ \(\left[ \begin{array}{l}4x-1=0\\x-2004=0\end{array} \right.\)

    ⇔ \(\left[ \begin{array}{l}x=1/4\\x=2004\end{array} \right.\)

    c) `(x+1)^2 = x+1`

    ⇔ `(x+1)^2 – x-1 =0`

    ⇔ `(x+1)^2 -(x+1)=0`

    ⇔ `(x+1)(x+1-1)=0`

    ⇔ `(x+1)x=0`

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

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

    0
    2021-08-09T16:08:08+00:00

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