i,2.|x-3|+4=6 j,27-3.|2x-4|=3 k,|2x+4|= x – 4 06/11/2021 Bởi aikhanh i,2.|x-3|+4=6 j,27-3.|2x-4|=3 k,|2x+4|= x – 4
`i. 2. |x-3|+4=6` `=> 2. |x-3|=6-4` `=> 2. |x-3|=2` `=> |x-3|=2: 2` `=> |x-3|=1` `=> x-3=1` hoặc `x-3=-1` $+)$ `x-3=1` `x=1+3` `x=4` $+)$ `x-3=-1` `x=-1+3` `x=2` Vậy `x \in {2; 4}.` `j. 27-3. |2x-4|=3` `=> 3. |2x-4|=27-3` `=> 3. |2x-4|=24` `=> |2x-4|=24: 3` `=> |2x-4|=8` `=> 2x-4=8` hoặc `2x-4=-8` $+)$ `2x-4=8` `2x=8+4` `2x=12` `x=12: 2` `x=6` $+)$ `2x-4=-8` `2x=-8+4` `2x=-4` `x=-4: 2` `x=-2` Vậy `x \in {-2; 6}.` `k. |2x+4|= x – 4` \(\left[ \begin{array}{l}2x+4=x-4\\2x+4=4-x\end{array} \right.\) \(\left[ \begin{array}{l}2x-x=-4-4\\2x+x=4-4\end{array} \right.\) \(\left[ \begin{array}{l}x=-8\\x=0\end{array} \right.\) Bình luận
i, 2 . | x – 3 | + 4 = 6 2 . | x – 3 | = 2 | x – 3 | = 1 `=>` \(\left[ \begin{array}{l}x-3=1\\x-3=-1\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\) Vậy x ∈ { 2 ; 4 } j, 27 – 3 . | 2x – 4 | = 3 3 . | 2x – 4 | = 9 | 2x – 4 | = 3 `=>` \(\left[ \begin{array}{l}2x-4=3\\2x-4=-3\end{array} \right.\) `=>` \(\left[ \begin{array}{l}2x=7\\2x=1\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=\frac{7}{2}\\x=\frac{1}{2}\end{array} \right.\) Vậy x ∈ { `7/2` ; `1/2` } k, | 2x + 4 | = x – 4 `=>` \(\left[ \begin{array}{l}2x+4=x-4\\2x+4=4-x\end{array} \right.\) `=>` \(\left[ \begin{array}{l}2x-x=-4-4\\2x+x=4-4\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=-8\\x=0\end{array} \right.\) Vậy x ∈ { -8 ; 0 } Bình luận
`i. 2. |x-3|+4=6`
`=> 2. |x-3|=6-4`
`=> 2. |x-3|=2`
`=> |x-3|=2: 2`
`=> |x-3|=1`
`=> x-3=1` hoặc `x-3=-1`
$+)$ `x-3=1`
`x=1+3`
`x=4`
$+)$ `x-3=-1`
`x=-1+3`
`x=2`
Vậy `x \in {2; 4}.`
`j. 27-3. |2x-4|=3`
`=> 3. |2x-4|=27-3`
`=> 3. |2x-4|=24`
`=> |2x-4|=24: 3`
`=> |2x-4|=8`
`=> 2x-4=8` hoặc `2x-4=-8`
$+)$ `2x-4=8`
`2x=8+4`
`2x=12`
`x=12: 2`
`x=6`
$+)$ `2x-4=-8`
`2x=-8+4`
`2x=-4`
`x=-4: 2`
`x=-2`
Vậy `x \in {-2; 6}.`
`k. |2x+4|= x – 4`
\(\left[ \begin{array}{l}2x+4=x-4\\2x+4=4-x\end{array} \right.\)
\(\left[ \begin{array}{l}2x-x=-4-4\\2x+x=4-4\end{array} \right.\)
\(\left[ \begin{array}{l}x=-8\\x=0\end{array} \right.\)
i, 2 . | x – 3 | + 4 = 6
2 . | x – 3 | = 2
| x – 3 | = 1
`=>` \(\left[ \begin{array}{l}x-3=1\\x-3=-1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\)
Vậy x ∈ { 2 ; 4 }
j, 27 – 3 . | 2x – 4 | = 3
3 . | 2x – 4 | = 9
| 2x – 4 | = 3
`=>` \(\left[ \begin{array}{l}2x-4=3\\2x-4=-3\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x=7\\2x=1\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\frac{7}{2}\\x=\frac{1}{2}\end{array} \right.\)
Vậy x ∈ { `7/2` ; `1/2` }
k, | 2x + 4 | = x – 4
`=>` \(\left[ \begin{array}{l}2x+4=x-4\\2x+4=4-x\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x-x=-4-4\\2x+x=4-4\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-8\\x=0\end{array} \right.\)
Vậy x ∈ { -8 ; 0 }