B1 a)\(\left(x-\dfrac{1}{2}\right)^3=8\) b)(x-5)^2=49 c)(2x-3)^3=125
\(\Rightarrow\)\(\left(x-\dfrac{1}{2}\right)^3=2^3\) \(\Rightarrow\)(x-5)^2=7^2 \(\Rightarrow\)(2x-3)^3=5^3
\(\Rightarrow\)x-\(\dfrac{1}{2}=2\) \(\Rightarrow\)x-5=7 \(\Rightarrow\)2x-3=5
\(\Rightarrow\)x=\(\dfrac{5}{2}\) \(\Rightarrow\)x=12 \(\Rightarrow\)2x=8
Vậy x=\(\dfrac{5}{2}\) Vậy=12 \(\Rightarrow\)x=4
Vậy x=4
d)\(\left(x-\dfrac{5}{4}\right)^3=8\) e)(2x-1)^2=25 f)\(\left(x-\dfrac{4}{3}\right)^5=1\)
\(\Rightarrow\)\(\left(x-\dfrac{5}{4}\right)^3=2^3\) \(\Rightarrow\)(2x-1)^2=5^2 \(\Rightarrow\left(x-\dfrac{4}{3}\right)^5=1^5\)
\(\Rightarrow\)\(x-\dfrac{5}{4}=2\) \(\Rightarrow\)2x-1=5 \(\Rightarrow\) \(x-\dfrac{4}{5}=1\)
\(\Rightarrow x=\dfrac{13}{4}\) \(\Rightarrow\)2x=6 \(\Rightarrow x=\dfrac{9}{5}\)
Vậy x=\(\dfrac{13}{4}\) \(\Rightarrow\) x=3 Vậy x=\(\dfrac{9}{5}\)
Vẫy x=3
B2 a)2\(^{2x+1}=8\) b) 3\(^x\).5=15 c)\(5^{x+1}.2=50\)
\(\Rightarrow\)\(2^{2x+1}=2^3\) \(\Rightarrow\) 3^x=3 \(\Rightarrow5^{x+1}=25\)
\(\Rightarrow2x+1=3\) \(\Rightarrow\) 3^x=3^1 \(\Rightarrow5^{x+1}=5^2\)
\(\Rightarrow2x=2\) \(\Rightarrow\) x=1 \(\Rightarrow x+1=2\)
\(\Rightarrow x=1\) Vậy x=1 \(\Rightarrow x=1\)
Vậy x=1 Vậy x=1
Bài 1:
a) Ta có: \(\left(x-\dfrac{1}{2}\right)^3=8\)
\(\Leftrightarrow x-\dfrac{1}{2}=2\)
hay \(x=\dfrac{5}{2}\)
b) Ta có: \(\left(x-5\right)^2=49\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
c) Ta có: \(\left(2x-3\right)^3=125\)
\(\Leftrightarrow2x-3=5\)
\(\Leftrightarrow2x=8\)
hay x=4
d) Ta có: \(\left(x-\dfrac{5}{4}\right)^3=8\)
\(\Leftrightarrow\left(x-\dfrac{5}{4}\right)=2\)
hay \(x=\dfrac{13}{4}\)
e) Ta có: \(\left(2x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
f) Ta có: \(\left(x-\dfrac{4}{3}\right)^5=1\)
\(\Leftrightarrow x-\dfrac{4}{3}=1\)
hay \(x=\dfrac{7}{3}\)
Bài 2:
a) Ta có: \(2^{2x+1}=8\)
\(\Leftrightarrow2x+1=3\)
\(\Leftrightarrow2x=2\)
hay x=1
b) Ta có: \(3^x\cdot5=15\)
\(\Leftrightarrow3^x=3\)
hay x=1
c) Ta có: \(5^{x+1}\cdot2=50\)
\(\Leftrightarrow5^{x+1}=25\)
\(\Leftrightarrow x+1=2\)
hay x=1
