a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
b)\(\left(2x-3\right)^3=343\)
\(\left(2x-3\right)^3=7^3\)
\(2x-3=7\)
\(2x=7+3\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
a) Ta có: \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
<=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
<=> \(x=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy x=5/6
b)\(\left(2x-3\right)^3=343\)
<=>\(2x-3=\sqrt[3]{343}=7\)
<=> 2x=10 <=> x=5
c) \(\left(\frac{1}{3}\right)^{2x}+1=\frac{1}{7}\)
<=>\(\left(\frac{1}{3}\right)^{2x}=\frac{-6}{7}\)
<=> \(\left(\frac{1}{3^x}\right)^2=-\frac{6}{7}\)(vô lí vì \(\left(\frac{1}{3^x}\right)^2\ge0\))
Vậy ko tìm được x thỏa mãn.
d)\(\left(2x-3\right)^2=9\)
=>\(\left[\begin{array}{nghiempt}2x-3=3\\2x-3=-3\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=3\\x=0\end{array}\right.\)
Vậy x=3 hoặc x=0.
e) \(\left(x-3\right)^6=\left(x-3\right)^7\)
<=> \(\left(x-3\right)^7-\left(x-3\right)^6=0\)
<=> \(\left(x-3\right)^6\left(x-3-1\right)=0\)
<=>\(\left(x-3\right)^6\left(x-4\right)=0\)
<=> \(\left[\begin{array}{nghiempt}x-3=0\\x-4=0\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=4\end{array}\right.\)
Vậy x \(\in\left\{3;4\right\}\)
d) \(\left(2x-3\right)^2=9\)
\(\left(2x-3\right)^2=3^2\)
\(2x-3=3\)
\(2x=3+3\)
\(2x=6\)
\(x=\frac{6}{2}\)
\(x=3\)
e) \(\left(x-3\right)^6=\left(x-3\right)^7\)
\(\Rightarrow x-3=1\)
\(x=1+3=4\)
