Tìm x: $(2x+1)(5x-1)= $$20x^{2}$ $-16x-1$ 10/07/2021 Bởi Ayla Tìm x: $(2x+1)(5x-1)= $$20x^{2}$ $-16x-1$
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}}` Bình luận
`Pt ⇔ 10x^2 – 2x + 5x – 1 = 20x^2 – 16x – 1` `⇔ 10x^2 – 19x = 0` `⇔ x(10x – 19) = 0` `⇔` \(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\) `⇔ \(\left[ \begin{array}{l}x=0\\10x=19\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=0\\x=19/10\end{array} \right.\) Bình luận
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}}`
`Pt ⇔ 10x^2 – 2x + 5x – 1 = 20x^2 – 16x – 1`
`⇔ 10x^2 – 19x = 0`
`⇔ x(10x – 19) = 0`
`⇔` \(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\)
`⇔ \(\left[ \begin{array}{l}x=0\\10x=19\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=19/10\end{array} \right.\)