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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x+1^3=2^5-\left(-1^3\right)\)
\(\Rightarrow x+1=33\)
=> x = 32
b) \(3^7-x=1^4-\left(-3^5\right)\)
\(\Rightarrow2187-x=1+243=244\)
=> x = 1943
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. (2x - 1)5 = x5
=> 2x - 1 = x
2x - x = 1
1x = 1
x = 1
Vậy .......
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: x<>-1/2
\(\dfrac{x-1}{2x+1}=\dfrac{2}{3}\)
=>\(2\left(2x+1\right)=3\left(x-1\right)\)
=>\(4x+2=3x-3\)
=>\(4x-3x=-3-2\)
=>x=-5(nhận)
b: ĐKXĐ: x<>1/2
\(\dfrac{x-2}{2x-1}=\dfrac{-1}{3}\)
=>\(3\left(x-2\right)=-1\left(2x-1\right)\)
=>\(3x-6=-2x+1\)
=>\(3x+2x=1+6\)
=>5x=7
=>x=7/5(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
`|2x+1|-3=x+4`
`<=>|2x+1|=x+4+3=x+7(x>=-7)`
`**2x+1=x+7`
`<=>x=7-1=6(tm)`
`**2x+1=-x-7`
`<=>3x=-6`
`<=>x=-2(tm)`
`|3x-5|=1-3x(x<=1/3)`
`**3x-5=1-3x`
`<=>6x=6`
`<=>x=1(l)`
`**3x-5=3x-1`
`<=>-5=-1` vô lý
`|2x+2|+|x-1|=10`
Nếu `x>=1`
`pt<=>2x+2+x-1=10`
`<=>3x+1=10`
`<=>3x=9`
`<=>x=3(tm)`
Nếu `x<=-1`
`pt<=>-2x-2+1-x=10`
`<=>-1-3x=10`
`<=>-11=3x`
`<=>x=-11/3(tm)`
Nếu `-1<=x<=1`
`pt<=>2x+2+1-x=10`
`<=>x+3=10`
`<=>x=7(l)`
Vậy `S={3,-11/3}`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
a) ta xét các trường hợp:
+ Với x \(\)<-1
\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow-x+4-x+3+x+1=5\)
\(\Rightarrow-x+8=5\)
\(\Rightarrow-x=-3\)
\(\Rightarrow x=3\)(không thỏa mãn )
+Với -1\(\le\)x<3
\(\)\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow-x+4-x+3-x-1=5\)
\(\Rightarrow-3x+6=5\)
\(\Rightarrow-3x=-1\)
\(\Rightarrow x=\frac{1}{3}\)(thỏa mãn)
+ Với 3\(\le\)x<4
\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow-x+4+x-3-x-1=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)(không thỏa mãn)
+ Với x\(\ge\)4
\(\Rightarrow\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)
\(\Rightarrow x-4+x-3-x-1=5\)
\(\Rightarrow x-8=5\)
\(\Rightarrow x=13\)(thỏa mãn)
Vậy \(x\in\left\{\frac{1}{3};13\right\}\)thì \(\left|x-4\right|+\left|x-3\right|-\left|x+1\right|=5\)