Tìm x: $(2x+1)(5x-1)= $$20x^{2}$ $-16x-1$

Question

Tìm x:
$(2x+1)(5x-1)= $$20x^{2}$ $-16x-1$

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Adela 9 months 2020-10-29T01:01:58+00:00 2 Answers 63 views 0

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    0
    2020-10-29T01:03:30+00:00

    Đáp án:

     

    Giải thích các bước giải:

     `(2x+1)(5x-1)=20x^2-16x-1`

    `⇔10x^2-2x+5x-1-20x^2+16x+1=0`

    `⇔-10x^2+19x=0`

    `⇔x(-10x+19)=0`

    `⇔`\(\left[ \begin{array}{l}x=0\\-10x+19=0\end{array} \right.\) 

    `⇔`\(\left[ \begin{array}{l}x=0\\-10x=-19\end{array} \right.\) 

    `⇔`\(\left[ \begin{array}{l}x=0\\x=\frac{19}{10}\end{array} \right.\)

    Vậy `x∈{0;{19}/{10}}`

    0
    2020-10-29T01:03:46+00:00

    `Pt ⇔ 10x^2 – 2x + 5x – 1 = 20x^2 – 16x – 1`

    `⇔ 10x^2 – 19x = 0`

    `⇔ x(10x – 19) = 0`

    `⇔` \(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\)

    `⇔ \(\left[ \begin{array}{l}x=0\\10x=19\end{array} \right.\) 

    `⇔` \(\left[ \begin{array}{l}x=0\\x=19/10\end{array} \right.\) 

     

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )