1.giải pt: a)(x-3) (2x+1) (x-4x+3) = 0 b) (3x-2)(2x+5)(8x-3)(4x+1)(3x-1)=0 10/10/2021 Bởi Emery 1.giải pt: a)(x-3) (2x+1) (x-4x+3) = 0 b) (3x-2)(2x+5)(8x-3)(4x+1)(3x-1)=0
Đáp án: Giải thích các bước giải: `a)(x-3) (2x+1) (x-4x+3) = 0` `↔(x-3)(2x+1)(3-3x)=0` `↔` \(\left[ \begin{array}{l}x-3=0\\2x+1=0\\3-3x=0\end{array} \right.\) `↔` \(\left[ \begin{array}{l}x=3\\x=-\dfrac{1}{2}\\x=1\end{array} \right.\) Vậy `S={-1/2;1;3}` `b)(3x-2)(2x+5)(8x-3)(4x+1)(3x-1)=0` `↔` \(\left[ \begin{array}{l}3x-2=0\\2x+5=0\\8x-3=0\\4x+1=0\\3x-1=0\end{array} \right.\) `↔` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{2}\\x=\dfrac{3}{8}\\x=-\dfrac{1}{4}\\x=\dfrac{1}{3}\end{array} \right.\) Vậy `S={-1/4;-5/2;2/3;1/3;3/8}` Bình luận
Đáp án: Giải thích các bước giải: `a)(x-3) (2x+1) (x-4x+3) = 0` `<=>`\(\left[ \begin{array}{l}x-3=0\\2x+1=0\\x-4x+3=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=3\\2x=-1\\-3x=-3\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=3\\x=-1/2\\x=1\end{array} \right.\) Vậy `S={3;-1/2;1}` `b) (3x-2)(2x+5)(8x-3)(4x+1)(3x-1)=0` `<=>`\(\left[ \begin{array}{l}3x-2=0\\2x+5=0\\8x-3=0\\4x+1=0\\3x-1=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}3x=2\\2x=-5\\8x=3\\4x=-1\\3x=1\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{-5}{2}\\x=\frac{3}{8}\\x=\frac{-1}{4}\\x=\frac{1}{3}\end{array} \right.\) Vậy `S= {2/3;-5/2;3/8;-1/4;1/3}` Bình luận
Đáp án:
Giải thích các bước giải:
`a)(x-3) (2x+1) (x-4x+3) = 0`
`↔(x-3)(2x+1)(3-3x)=0`
`↔` \(\left[ \begin{array}{l}x-3=0\\2x+1=0\\3-3x=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=3\\x=-\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `S={-1/2;1;3}`
`b)(3x-2)(2x+5)(8x-3)(4x+1)(3x-1)=0`
`↔` \(\left[ \begin{array}{l}3x-2=0\\2x+5=0\\8x-3=0\\4x+1=0\\3x-1=0\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{2}\\x=\dfrac{3}{8}\\x=-\dfrac{1}{4}\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={-1/4;-5/2;2/3;1/3;3/8}`
Đáp án:
Giải thích các bước giải:
`a)(x-3) (2x+1) (x-4x+3) = 0`
`<=>`\(\left[ \begin{array}{l}x-3=0\\2x+1=0\\x-4x+3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\2x=-1\\-3x=-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=-1/2\\x=1\end{array} \right.\)
Vậy `S={3;-1/2;1}`
`b) (3x-2)(2x+5)(8x-3)(4x+1)(3x-1)=0`
`<=>`\(\left[ \begin{array}{l}3x-2=0\\2x+5=0\\8x-3=0\\4x+1=0\\3x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}3x=2\\2x=-5\\8x=3\\4x=-1\\3x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\frac{2}{3}\\x=\frac{-5}{2}\\x=\frac{3}{8}\\x=\frac{-1}{4}\\x=\frac{1}{3}\end{array} \right.\)
Vậy `S= {2/3;-5/2;3/8;-1/4;1/3}`