a) \(\left(2x+3\right)^3=27\Rightarrow\left(2x+3\right)^3=3^3\Rightarrow2x+3=3\Rightarrow x=0\)
b) \(\left(x+3\right)^3=125\Rightarrow\left(x+3\right)^3=5^3\Rightarrow x+3=5\Rightarrow x=2\)
c) \(\left(x+1\right)^4=\left(2x\right)^4\Rightarrow x+1=2x\Rightarrow x=1\)
d) \(\left(2x-1\right)^5=x^5\Rightarrow2x-1=x\Rightarrow x=1\)
\(g,\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
\(2x+1=3\)
\(2x=3-1\)
\(2x=2\)
\(x=2:2\)
\(x=1\)
Vậy \(x=1\)
\(h,\left(x-2\right)^3=64\)
\(\left(x-2\right)^3=4^3\)
\(x-2=4\)
\(x=4+2\)
\(x=6\)
Vậy \(x=6\)
Bài 15:
\(a,\left(2x+3\right)^3=27\)
\(\left(2x+3\right)^3=3^3\)
\(2x+3=3\)
\(2x=3-3\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
\(b,\left(x+3\right)^3=125\)
\(\left(x+3\right)^3=5^3\)
\(x+3=5\)
\(x=5-3\)
\(x=2\)
Vậy \(x=2\)
\(c,\left(x+1\right)^4=\left(2x\right)^4\)
\(x+1=2x\)
\(2x-x=1\)
\(\left(2-1\right)x=1\)
\(x=1\)
Vậy \(x=1\)
\(d,\left(2x-1\right)^5=x^5\)
\(2x-1=x\)
\(2x-x=1\)
\(x=1\)
Vậy \(x=1\)
