Tìm x , biết : a) / 4x-4/ + /x-2/=5 b) /x+3/+/x+4/=8 ( / là giá trị tuyệt đối) 20/07/2021 Bởi Mary Tìm x , biết : a) / 4x-4/ + /x-2/=5 b) /x+3/+/x+4/=8 ( / là giá trị tuyệt đối)
Đáp án: b. \(\left[ \begin{array}{l}x = \dfrac{1}{2}\\x = – \dfrac{{15}}{2}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.\left| {4x – 4} \right| + \left| {x – 2} \right| = 5\\ \to \left[ \begin{array}{l}4x – 4 + x – 2 = 5\left( {DK:x \ge 2} \right)\\4x – 4 – x + 2 = 5\left( {DK:2 > x \ge 1} \right)\\4x – 4 + x – 2 = – 5\left( {DK:1 > x} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}5x = 11\\3x = 7\\5x = 1\end{array} \right. \to \left[ \begin{array}{l}x = \dfrac{{11}}{5}\left( {TM} \right)\\x = \dfrac{7}{3}\left( l \right)\\x = \dfrac{1}{5}\left( {TM} \right)\end{array} \right.\\b.\left| {x + 3} \right| + \left| {x + 4} \right| = 8\\ \to \left[ \begin{array}{l}x + 3 + x + 4 = 8\left( {DK:x \ge – 3} \right)\\ – x – 3 + x + 4 = 8\left( {DK: – 3 > x \ge – 4} \right)\\x + 3 + x + 4 = – 8\left( {DK: – 4 > x} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}2x = 1\\1 = 8\left( l \right)\\2x = – 15\end{array} \right. \to \left[ \begin{array}{l}x = \dfrac{1}{2}\\x = – \dfrac{{15}}{2}\end{array} \right.\left( {TM} \right)\end{array}\) Bình luận
Lm câu a thoii đc chớ ak? a, `|4x-4|+|x-2|=5` `=> 4|x-1|+|x-2|=5` (1) Ker bảng : x | 1 2 | x – 1 | – 0 + | + | x – 2 | – | – 0 + |Th1 : x < 1 => (1) `⇔ 4 (1-x)+(2-x)=5` `⇔4-4x+2-x=5` `⇔6-5x=5` `⇔5x=1` `⇔x=1/5` (tm x <1) Th2 : `1≤x≤2` `=> (1) ⇔ 4x-4+2-x=5` `⇔3x-2=5` `⇔3x=7` `⇔x=7/3` (ktmđk) Th2 : x >2 `=> (1) ⇔ 4x-4+x-2=5` `=> 5x – 6 = 5` `=> 5x = 11` `=> x = 11/5` ™Vậy `x∈{11/5 ; 1/5}` Bình luận
Đáp án:
b. \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = – \dfrac{{15}}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {4x – 4} \right| + \left| {x – 2} \right| = 5\\
\to \left[ \begin{array}{l}
4x – 4 + x – 2 = 5\left( {DK:x \ge 2} \right)\\
4x – 4 – x + 2 = 5\left( {DK:2 > x \ge 1} \right)\\
4x – 4 + x – 2 = – 5\left( {DK:1 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = 11\\
3x = 7\\
5x = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{11}}{5}\left( {TM} \right)\\
x = \dfrac{7}{3}\left( l \right)\\
x = \dfrac{1}{5}\left( {TM} \right)
\end{array} \right.\\
b.\left| {x + 3} \right| + \left| {x + 4} \right| = 8\\
\to \left[ \begin{array}{l}
x + 3 + x + 4 = 8\left( {DK:x \ge – 3} \right)\\
– x – 3 + x + 4 = 8\left( {DK: – 3 > x \ge – 4} \right)\\
x + 3 + x + 4 = – 8\left( {DK: – 4 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 1\\
1 = 8\left( l \right)\\
2x = – 15
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = – \dfrac{{15}}{2}
\end{array} \right.\left( {TM} \right)
\end{array}\)
Lm câu a thoii đc chớ ak?
a,
`|4x-4|+|x-2|=5`
`=> 4|x-1|+|x-2|=5` (1)
Ker bảng :
x | 1 2 |
x – 1 | – 0 + | + |
x – 2 | – | – 0 + |
Th1 : x < 1
=> (1) `⇔ 4 (1-x)+(2-x)=5`
`⇔4-4x+2-x=5`
`⇔6-5x=5`
`⇔5x=1`
`⇔x=1/5` (tm x <1)
Th2 : `1≤x≤2`
`=> (1) ⇔ 4x-4+2-x=5`
`⇔3x-2=5`
`⇔3x=7`
`⇔x=7/3` (ktmđk)
Th2 : x >2
`=> (1) ⇔ 4x-4+x-2=5`
`=> 5x – 6 = 5`
`=> 5x = 11`
`=> x = 11/5` ™
Vậy `x∈{11/5 ; 1/5}`