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1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
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b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
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(4x - 9) (2,5 + 2/3x)=0
=> 4x-9 = 0 hoặc 2,5 +2/3x = 0
=> 4x = 9 hoặc 2/3x = -2,5
=> x = 9/4 hoặc x = -7,5/2
kết luận : vậy x thuộc {9/4; -7,5/2}
(x - 5)2 = ( 1 - 3x)2
=> x-5 = 1-3x
=> x-5+3x = 1
=>4x-5 =1
=> 4x=6
=> x=3/2
|x|=3
=> X=3 hoặc x=-3
3| x+1| - 2=1
=> 3lx+1l = 3
=> lx+1l =1
=> x+1 = 1 hoặc x+1= -1
=> x=0 hoặc x = -2
3|x + 1| + 2=1
=> 3lx+1l = -1
=> lx+1l = -1/3
vô lý vì giá trị tuyệt đối của 1 số luôn luôn lớn hơn hoặc bằng 0
=> x thuộc rỗng
![](https://rs.olm.vn/images/avt/0.png?1311)
a, +) Xét \(x\ge2,5\) có:
\(x-1,5+x-2,5=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( không t/m )
+) Xét \(1,5\le x< 2,5\) có:
\(x-1,5+2,5-x=0\)
\(\Leftrightarrow1=0\) ( ko t/m )
+) Xét x < 1,5 có:
\(1,5-x+2,5-x=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\) ( ko t/m )
Vậy không có giá trị x thỏa mãn
b, \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{matrix}\right.\Leftrightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Leftrightarrow3x\ge0\Leftrightarrow x\ge0\)
\(\Leftrightarrow x+3+x+1=3x\)
\(\Leftrightarrow x=4\)
Vậy x = 4
c, \(\left|x-7\right|=1-2x\)
+) Xét \(x\ge7\) có:
\(x-7=1-2x\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)( ko t/m )
+) Xét x < 7 có:
\(7-x=1-2x\Leftrightarrow x=-6\) ( t/m )
Vậy x = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : |x - 1,5| + |2,5 - x| \(\ge\left|x-1,5+2,5-x\right|\)
<=> |x - 1,5| + |2,5 - x| \(\ge\left|1\right|\)
=> |x - 1,5| + |2,5 - x| \(\ge1\)
Vậy GTNN của biểu thức là : 1
Khi 1,5 \(\le x\le2,5\)
Vậy nên đề sai nhá
c) \(\left|x-7\right|=1-2x\)
khi \(x\ge\frac{1}{2}\), biểu thức có dạng:
\(\orbr{\begin{cases}x-7=1-2x\\x-7=2x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=8\\-x=6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-6\end{cases}}}\)
8/3 (nhận); -6 (loại)
vậy x=8/3
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\(\left(\frac{2}{3}x-\frac{4}{9}\right)\left[\frac{1}{2}+\left(-\frac{3}{7}:x\right)\right]=0\)
\(\Rightarrow\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}.\frac{1}{x}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}.\frac{1}{x}=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{1}{x}=\frac{7}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}}\)
Vậy x = 2/3 , x = 6/7
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\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{7}\div x\right)=0\)
\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{8}.\frac{1}{x}\right)=0\)
\(\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}.x=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{1}{x}=\frac{7}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
\(\left|x-2,5\right|-\frac{3}{4}=0\)
\(\Leftrightarrow\left|x-\frac{1}{4}\right|=\frac{3}{4}\)
\(\Leftrightarrow x-\frac{1}{4}=\pm\frac{3}{4}\)
TH1:\(x-\frac{1}{4}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{4}=1\)
TH2:\(x-\frac{1}{4}=-\frac{3}{4}\)
\(\Leftrightarrow x=-\frac{3}{4}+\frac{1}{4}=-\frac{1}{2}\)
Vậy \(x=1;-\frac{1}{2}\)