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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
\(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=0\end{matrix}\right.\)
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2(x - 1) - 3(2x + 2) - 4(2x + 3) = 16
2x - 2 - 6x + 6 - 8x + 12 = 16
(2x - 6x - 8x) - (2 + 6 + 12) = 16
(-12) . x - 20 =16
(-12) . x = 16 + 20 = 36
x = 36 : (-12) = -3
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2 (x+1) -3 (2x+2) -4(2x+3)=16
2x+2-6x-6-8x-12=16
2x-6x-8x=16-2+6+12
-12x=32
x=32/-12
x=-2
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\(\Leftrightarrow x\cdot\dfrac{1}{2}=\dfrac{7}{16}-\dfrac{1}{8}=\dfrac{5}{16}\)
hay \(x=\dfrac{5}{8}\)
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Ta có: \(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Rightarrow2x-2-6x-6-8x-12=16\)
\(\Rightarrow2x-6x-8x=16+2+6+12\)
\(\Rightarrow-12x=36\)
\(\Rightarrow x=-3\)
Vậy x = -3
2x-1*1-2x=-16
2x-1-2x=-16
2x-2x-1=-16
0-1=-16
suy ra ko có x thỏa mãn