a)=>\(\left(2x+1\right)^2=\frac{1}{9}\)
\(=>\left(2x+1\right)^2=\frac{1}{3^2}\)
\(=>2x+1=\frac{1}{3}\)
\(=>2x=\frac{1}{3}-1\)
\(=>2x=\frac{-2}{3}\)
\(=>x=\frac{-2}{3}:2\)
\(=>x=\frac{-1}{3}\)
Vậy x = \(-\frac{1}{3}\)
b)\(=>\left(x-2\right)^3=27\)
\(=\left(x-2\right)^3=3^3\)
\(=>x-2=3\)
\(=>x=3+2\)
\(=>x=5\)
Vậy x = 5
c)=>x.x-x=0
TH1:\(\hept{\begin{cases}x.x=0\\x=0\end{cases}}\)\(=>\hept{\begin{cases}x=0\\x=0\end{cases}}\)
TH2:\(x.x=1.x=>x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d)\(x^4=27.x\)
\(=>x^4-27x=0\)
\(=>x^4-\left[\left(3\right)^3.x\right]=0\)
\(=>x^3.x-3^3.x=0\)
\(=>x.\left(x^3-3^3\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x^3-3^3=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x^3=3^3\end{cases}=>\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
X khong thể bằng (-3) được
Vậy x \(\in\){0;3}
a) Ta có: \(\left(2x+1\right)^2-\frac{1}{9}=0\)
\(\left(2x+1\right)^2=\frac{1}{9}\)
mà \(\frac{1}{9}=\left(\frac{1}{3}\right)^2=\left(-\frac{1}{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{3}\\2x+1=-\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-\frac{2}{3}\\2x=-\frac{4}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{3};-\frac{2}{3}\right\}\)
b) (x-2)3 + 27 = 0
(x-2)3 = -27
mà -27=(-3)3
=> x-2=-3
=> x= -1
c)Ta có: x2 - x = 0
x . (x-1) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(x\in\left\{0;1\right\}\)
d) x4 = 27x
x4 : x = 27
x3 = 27
mà 27 = 33
=> x= 3
