(2x-1)^4=16
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a: =>3|x-4|=16-2x
TH1: x>=4
=>3x-12=16-2x
=>5x=28
=>x=28/5(nhận)
TH2: x<4
=>12-3x=16-2x
=>-x=4
=>x=-4(nhận)
b: =>|2x-5|=7+4x+4=4x+11
TH1: x>=5/2
=>4x+11=2x-5
=>2x=-16
=>x=-8(loại)
TH2: x<5/2
=>4x+1=5-2x
=>6x=4
=>x=2/3(nhận)
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a) \(\left(2x-1\right)^4=16\)
\(\)TH1: \(\left(2x-1\right)^4=2^4\)
\(=>2x-1=2\)
\(2x=2+1\)
\(2x=3\)
\(x=\dfrac{3}{2}\)
TH2: \(\left(2x-1\right)^4=\left(-2\right)^4\)
\(=>2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=\dfrac{-1}{2}\)
Vậy x = \(\dfrac{3}{2}\) hoặc x = \(\dfrac{-1}{2}\)
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b) \(\left(2x+1\right)^3=125\) ( mình nghĩ đề bài đúng là vầy )
\(\left(2x+1\right)^3=5^3\)
\(=>2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
Vậy x = \(2\)
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b: =>2x=16/2=8
=>x=4
a: Sửa đề: (3/2)^2x-1=(3/2)^5x-4
=>2x-1=5x-4
=>-3x=-3
=>x=1
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a:
ĐKXĐ: x<>-1/2
Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì
\(2x^3+x^2+2x+1+1⋮2x+1\)
=>\(2x+1\inƯ\left(1\right)\)
=>2x+1 thuộc {1;-1}
=>x thuộc {0;-1}
b:
ĐKXĐ: x<>1/3
\(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)
=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1
=>2 chia hết cho 3x-1
=>3x-1 thuộc {1;-1;2;-2}
=>x thuộc {2/3;0;1;-1/3}
mà x nguyên
nên x thuộc {0;1}
c:
ĐKXĐ: x<>2
\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)
=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)
=>\(x+2⋮x-2\)
=>x-2+4 chia hết cho x-2
=>4 chia hết cho x-2
=>x-2 thuộc {1;-1;2;-2;4;-4}
=>x thuộc {3;1;4;0;6;-2}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x-1)4 = 16
=> (2x-1)4 = 24 hoặc (-2)4
=>\(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2+1\\2x=-2+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) (2x+1)4 = (2x+1)6
=> (2x+1)4 = (2x+1)4+2
=> (2x+1)4 = (2x+1)4 . (2x+1)2
=> (2x+1)4 - (2x+1)4 . (2x+1)2 = 0
=> (2x+1)4 . [1 - (2x+1)2] = 0
\(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=0\end{matrix}\right.\)
a) (2x-1)4 = 16
=> (2x-1)4 = 24 hoặc (-2)4
=>[2x−1=22x−1=−2⇒[2x=2+12x=−2+1⇒⎡⎢ ⎢⎣x=32x=−12
(2x-1)^4=16
2x-1=2
2x=3
x=3/2