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Bài 1:\(\left|2x-1\right|=2x-1\) khi \(x>0\)
b)\(\left|0,5-3x\right|=3x-0.5\) khi x= 4
c)\(\left|5x+1\right|-10x=0,5\) khi x= 0,1
Bài 2:Min A=0
Min B=-2
Bài 1:
a, \(\left|2x-1\right|=2x-1\)
+) Xét \(x\ge\dfrac{1}{2}\) ta có:
\(2x-1=2x-1\)
\(\Rightarrow x\) tùy ý với \(x\ge\dfrac{1}{2}\)
+) Xét \(x< \dfrac{1}{2}\) ta có:
\(1-2x=2x-1\)
\(\Rightarrow4x=2\)
\(\Rightarrow x=\dfrac{1}{2}\) ( không t/m )
Vậy...
b, \(\left|0,5-3x\right|=3x-0,5\)
+) Xét \(x\ge\dfrac{1}{6}\) ta có:
\(0,5-3x=3x-0,5\)
\(\Rightarrow6x=1\)
\(\Rightarrow x=\dfrac{1}{6}\) ( t/m )
+) Xét \(x< \dfrac{1}{6}\) ta có:
\(3x-0,5=3x-0,5\)
\(\Rightarrow x\) tùy ý với \(x< \dfrac{1}{6}\)
Vậy \(x\le\dfrac{1}{6}\)
c, \(\left|5x+1\right|-10x=0,5\)
+) Xét \(x\ge\dfrac{-1}{5}\) ta có:
\(5x+1-10x=0,5\)
\(\Rightarrow-5x=-0,5\)
\(\Rightarrow x=\dfrac{1}{10}\) ( t/m )
+) Xét \(x< \dfrac{-1}{5}\) ta có:
\(-5x-1-10x=0,5\)
\(\Rightarrow-15x=1,5\)
\(\Rightarrow x=\dfrac{-1}{10}\) ( không t/m )
Vậy \(x=\dfrac{1}{10}\)
Bài 2:
a, Ta có: \(-\left|x-3,5\right|\le0\)
\(\Rightarrow A=0,5-\left|x-3,5\right|\le3,5\)
Dấu " = " xảy ra khi \(-\left|x-3,5\right|=0\Rightarrow x=3,5\)
Vậy \(MIN_A=0,5\) khi x = 3,5
b, Ta có: \(-\left|1,4-x\right|\le0\)
\(\Rightarrow B=-\left|1,4-x\right|-2\le-2\)
Dấu " = " xảy ra khi \(-\left|1,4-x\right|=0\Rightarrow x=1,4\)
Vậy \(MIN_B=-2\) khi \(x=1,4\)
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a)\(\left|x-1\right|+3x=1\)
\(\Rightarrow\left|x-1\right|=1-3x\)
\(\Rightarrow\orbr{\begin{cases}x-1=1-3x\\x-1=1+3x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+\left(-1\right)=1+\left(-3x\right)\\x+\left(-1\right)=1+3x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-\left(-3x\right)=1-\left(-1\right)\\x-3x=1-\left(-1\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=2\\-2x=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
A ) | x - 1 | + 3x = 1
x - 1 + 3x = 1
( x + 3x ) - 1 = 1
4x - 1 = 1
4x = 1 + 1
4x = 2
x = 2 : 4
x = 0,5
Vậy x = 0,5
B) | 5x - 3 | - x = 7
5x - 3 - x = 7
( 5x - x ) - 3 = 7
4x - 3 = 7
4x = 7 + 3
4x = 10
x = 10 : 4
x = 2,5
Vậy x = 2,5
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\(a.12-\left|x-3\right|=5x+\)\(8\)
\(\Leftrightarrow\left|x-3\right|=5x+8-12\)
\(\Leftrightarrow\left|x-3\right|=5x-4\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=5x-4\\x-3=-5x+4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\frac{7}{6}\end{cases}}}\)
\(b.3-\frac{0.2x}{5}=\frac{7}{15}+1,4x\)
\(\Leftrightarrow\frac{15-0,2x}{5}=\frac{7+21x}{15}\)
\(\Leftrightarrow15.\left(15-0,2x\right)=5.\left(7+21x\right)\)
\(\Leftrightarrow225-3x=35+105x\)
\(\Leftrightarrow-3x-105x=35-225\)
\(\Leftrightarrow-108x=-190\)
\(\Rightarrow x=\frac{95}{54}\)
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a) |2x - 1| = 3
=> 2x - 1 = 3 hoặc 2x - 1 = -3
=> 2x = 4 hoặc 2x = -2
=> x = 2 hoặc x = -1
b) |2x + 5| + 5x = 9
|2x + 5| = 9 - 5x
=> 9 - 5x = 2x + 5 hoặc 9 - 5x = -(2x + 5)
=> 9 - 5 = 2x + 5x hoặc 9 - 5x = -2x - 5
=> 7x = 4 hoặc 9 + 5 = -2x + 5x
=> x = 4/7 hoặc 14 = 3x
=> x = 4/7 hoặc x = 14/3
\(\left(a\right)\left|2x-1\right|=3\)
\(\Rightarrow\hept{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Rightarrow\hept{\begin{cases}2x=4\\2x=-2\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-1\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
\(\left|1-2x\right|+x+2=0\)
⇒ \(\left|1-2x\right|+x=0-2\)
⇒ \(\left|1-2x\right|+x=-2\)
⇒ \(\left|1-2x\right|=-2-x\)
⇒ \(\left[{}\begin{matrix}1-2x=-2-x\\1-2x=2+x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}1+2=-x+2x\\1-2=x+2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3=1x\\-1=3x\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=3:1\\x=\left(-1\right):3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=3\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{3;-\frac{1}{3}\right\}.\)
Bài 4:
\(\left|5x-3\right|=\left|7-x\right|\)
⇒ \(\left[{}\begin{matrix}5x-3=7-x\\5x-3=x-7\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x+x=7+3\\5x-x=\left(-7\right)+3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}6x=10\\4x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=10:6\\x=\left(-4\right):4\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{5}{3}\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{3};-1\right\}.\)
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\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\)
\(=\dfrac{3+2+1}{6}+\dfrac{14+25-4}{35}=1+1=2\)