a, \(\left(x+1\right)^3:\left(x+1\right)=4\)
\(\Leftrightarrow\left(x+1\right)^2=4\)
\(\Leftrightarrow\left(x+1\right)^2=2^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy .....
b, \(3^x+60=87\)
\(\Leftrightarrow3^x=87-60\)
\(\Leftrightarrow3^x=27\)
\(\Leftrightarrow3^x=3^3\)
\(\Leftrightarrow x=3\)
Vậy ...
c, (Viết cái đề khó nhìn quá à)
d, \(2^x-64=2^6\)
\(\Leftrightarrow2^x-64=64\)
\(\Leftrightarrow2^x=64+64\)
\(\Leftrightarrow2^x=128\)
\(\Leftrightarrow2^x=2^7\)
\(\Leftrightarrow x=7\)
Vậy ...
a)
\(\left(x+1\right)^3:\left(x+1\right)=4\)
\(\left(x+1\right)^3:\left(x+1\right)^1=4\)
\(\left(x+1\right)^{3-1}=4\)
\(\left(x+1\right)^2=4\)
\(\left(x+1\right)^2=2^2\)
⇒ \(x+1=2\)
\(x=2-1=1\)
b)
\(3^x+60=87\)
\(3^x=87-60\)
\(3^x=27\)
\(3^x=3^3\) \(\Rightarrow x=3\)
c)
\(6\cdot2^x+3\cdot2^x=9\cdot29\)
\(2^x\cdot\left(6+3\right)=9\cdot29\)
\(2^x\cdot9=9\cdot29\)
\(\Rightarrow2^x=29\) \(\Rightarrow\) Không có x thỏa mãn điều kiện
d)
\(2^x-64=2^6\)
\(2^x-64=64\)
\(2^x=64-64\)
\(2^x=0\) \(\Rightarrow\) Không có x thỏa mãn điều kiện
