Tìm x :
7,5 -3. | 5-2x | = 1/4
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Q
0
Q
1
18 tháng 8 2021
a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)
\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)
\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)
18 tháng 8 2021
em muốn hỏi là tại sao 3,5 bên trên xuống dưới lại là 3 và -x +2/5 của em xuống dưới lại chuyển thành x-2/5 ạ mong anh giải đáp
PT
4
LT
1
5 tháng 2 2016
7,5 - 3 |5 - 2x| = -4,5 => -3 |5 - 2x| = -12 => |5 - 2x| = 4
=> 5 - 2x = 4 hoặc -4
=> -2x = -1 hoặc -9
=> x = 0,5 hoặc x = 4,5
MV
2
4 tháng 8 2016
3/5-2x/=7,5-(-4,5) \(\Rightarrow\)5-2x=4 hoặc 5-2x=-4 Vậy x=0.5 hoặc 4.5
\(\Rightarrow\)3/5-2x/=7,5+4,5 \(\Rightarrow\) 2x=5-4 hoặc2x=5-(-4) Nhớ tích nha!!!!!
\(\Rightarrow\)3/5-2x/=12 \(\Rightarrow\)2x=1 hoặc 2x=9
\(\Rightarrow\)/5-2x/=12:3 \(\Rightarrow\)x=1:2 hoặc x=9:2
\(\Rightarrow\)/5-2x/=4 \(\Rightarrow\)x=0.5 hoặc x=4.5
BS
4 tháng 8 2016
\(3\left|5-2x\right|=7,5+4,5\)
\(3\left|5-2x\right|=12\)
\(\left|5-2x\right|=12\div3=4\)
\(\Rightarrow5-2x=4\)hoặc \(5-2x=-4\)
\(5-2x=4\Rightarrow2x=5-4=1\Rightarrow x=\frac{1}{2}\)
\(5-2x=-4\Rightarrow2x=5+4=9\Rightarrow x=\frac{9}{2}=4,5\)
1 tháng 10 2021
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
ND
1
\(7,5-3\left|5-2x\right|=\frac{1}{4}\)
\(\Rightarrow7,5-3\left|5-2x\right|=0,25\)
\(\Rightarrow3\left|5-2x\right|=7,5-0,25=7,25\)
\(\Rightarrow\left|5-3x\right|=7,25:3=2,42\)
Từ đây tự làm nốt nhé!
\(7,5-3.\left|5-2x\right|=\frac{1}{4}\)
\(\Rightarrow3.\left|5-2x\right|=7.5-\frac{1}{4}\)
\(\Rightarrow3.\left|5-2x\right|=\frac{29}{4}\)
\(\Rightarrow\)\(\left|5-2x\right|=\frac{29}{4}:3\)
\(\Rightarrow\left|5-2x\right|=\frac{29}{12}\)
\(\Rightarrow\hept{\begin{cases}TH1,5-2x=\frac{29}{12}\Rightarrow2x=\frac{31}{12}\Rightarrow x=\frac{31}{24}\\TH2,5-2x=\frac{-29}{12}\Rightarrow2x=\frac{89}{12}\Rightarrow x=\frac{89}{24}\end{cases}}\)
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