\(a,\rightarrow6x^3=48\\ \rightarrow x^3=8\\ x^3=2^3\\ \rightarrow x=2\\ b,\left(x-1\right)^2=2^2hoặc\left(x-1\right)^2=\left(-2\right)^2\\\rightarrow x-1=2hoặcx-1=-2\\ TH1:x-1=2\\ \rightarrow x=3\\ TH2:x-1=-2\\ \rightarrow x=-1\)
\(c,\left(2x+1\right)^3=3^3\\ \rightarrow2x+1=3\\ \rightarrow2x=2\\ \rightarrow x=1\)
a, \(6x^3=48\Leftrightarrow2x^3=16\Leftrightarrow x^3=8\Leftrightarrow x=2\)
b, \(\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c, \(2x+1=3\Leftrightarrow x=1\)
d, \(\left[{}\begin{matrix}x+1=2x\\x+1=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
e, \(x^5-x^3=x^3\left(x^2-1\right)=0\Leftrightarrow x=0;x=-1;x=1\)
f, Ta có \(\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\)
mà \(\left(x+1,5\right)^2+\left(y-2,5\right)^2\ge0\)
Dấu ''='' xảy ra khi x = -1,5 ; y = 2,5
`a)`
`6x^3 - 8=40`
`=>6x^3=40+8`
`=> 6x^3=48`
`=>x^3=48:6`
`=>x^3=8`
`=>x=2`
`b)`
`(x-1)^2=4`
`=>(x-1)^2=2^2`
\(\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2+1\\x=-2+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
`c)`
`(2x+1)^3=27`
`=(2x+1)^3= 3^3`
`=>2x+1=3`
`=>2x=3-1`
`=>2x=2`
`=>x=2:2`
`=>x=1`
`d)`
`(x+1)^4 =(2x)^4`
`=>x+1=2x`
`=>x+1-2x=0`
`=>(x-2x+1)=0`
`=>(x-1)^2=0`
`=>(x-1)=0^2`
`=>x-1=0`
`=>x=0+1`
`=>x=1`
