Bài 6 a) |x – 5| = 12 b) |3x -12| = x+2 c) |3x + 12| = x+2 Giúp mik với ạ mai mik pk nộp rùi :((( 04/11/2021 Bởi Parker Bài 6 a) |x – 5| = 12 b) |3x -12| = x+2 c) |3x + 12| = x+2 Giúp mik với ạ mai mik pk nộp rùi :(((
Đáp án: Giải thích các bước giải: a) |x – 5| = 12 ⇔\(\left[ \begin{array}{l}x – 5 = 12\\x – 5 = -12\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x = 12+5\\x = -12+5\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x = 17\\x = -7\end{array} \right.\) Vậy x ∈ {17;-7} b) |3x -12| = x+2 ⇔\(\left[ \begin{array}{l}3x -12 = x+2\\3x -12 = -(x+2)\end{array} \right.\) ⇔\(\left[ \begin{array}{l}3x -12 = x+2\\3x -12 = -x-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}3x – x=2+12\\3x +x=-2+12\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x=14\\4x=10\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=7\\x=5/2\end{array} \right.\) Vậy x ∈ {7;5/2} c) |3x + 12| = x+2 ⇔\(\left[ \begin{array}{l}3x +12 = x+2\\3x +12 = -(x+2)\end{array} \right.\) ⇔\(\left[ \begin{array}{l}3x +12 = x+2\\3x +12 = -x-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}3x – x=2-12\\3x +x=-2-12\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x=-10\\4x=-14\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-5\\x=-7/2\end{array} \right.\) (không thỏa mãn) Vậy x ∈ ∅ Bình luận
`a)` `|x-5|=12` `<=>` \(\left[ \begin{array}{l}x-5=12\\x-5=-12\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=12+5\\x=-12+5\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=17\\x=-7\end{array} \right.\) Vậy `x∈{17;-7}` `b)` `|3x-12|=x+2` `<=>` \(\left[ \begin{array}{l}3x-12=x+2\\3x-12=-(x+2)\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x-x=2+12\\3x-12=-x-2\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x=14\\4x=10\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=7\\x=\dfrac{2}{5}\end{array} \right.\) Vậy `x∈{7;2/5}` `c)` `|3x+12|=x+2` `<=>` \(\left[ \begin{array}{l}3x+12=x+2\\3x+12=-(x+2)\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x-x=2-12\\3x+12=-x-2\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=-10\\4x=-14\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-5\\x=-\dfrac{7}{2}\end{array} \right.\) Vậy `x∈{-5;-7/2}` Bình luận
Đáp án:
Giải thích các bước giải:
a) |x – 5| = 12
⇔\(\left[ \begin{array}{l}x – 5 = 12\\x – 5 = -12\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x = 12+5\\x = -12+5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x = 17\\x = -7\end{array} \right.\)
Vậy x ∈ {17;-7}
b) |3x -12| = x+2
⇔\(\left[ \begin{array}{l}3x -12 = x+2\\3x -12 = -(x+2)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x -12 = x+2\\3x -12 = -x-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x – x=2+12\\3x +x=-2+12\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=14\\4x=10\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=7\\x=5/2\end{array} \right.\)
Vậy x ∈ {7;5/2}
c) |3x + 12| = x+2
⇔\(\left[ \begin{array}{l}3x +12 = x+2\\3x +12 = -(x+2)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x +12 = x+2\\3x +12 = -x-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x – x=2-12\\3x +x=-2-12\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2x=-10\\4x=-14\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-5\\x=-7/2\end{array} \right.\) (không thỏa mãn)
Vậy x ∈ ∅
`a)` `|x-5|=12`
`<=>` \(\left[ \begin{array}{l}x-5=12\\x-5=-12\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=12+5\\x=-12+5\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=17\\x=-7\end{array} \right.\)
Vậy `x∈{17;-7}`
`b)` `|3x-12|=x+2`
`<=>` \(\left[ \begin{array}{l}3x-12=x+2\\3x-12=-(x+2)\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x-x=2+12\\3x-12=-x-2\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}2x=14\\4x=10\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=7\\x=\dfrac{2}{5}\end{array} \right.\)
Vậy `x∈{7;2/5}`
`c)` `|3x+12|=x+2`
`<=>` \(\left[ \begin{array}{l}3x+12=x+2\\3x+12=-(x+2)\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}3x-x=2-12\\3x+12=-x-2\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=-10\\4x=-14\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-5\\x=-\dfrac{7}{2}\end{array} \right.\)
Vậy `x∈{-5;-7/2}`