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\(2^x.4=128\)
\(\Rightarrow2^x=128:4\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
tíc mình nha
Ta có: A = 1 + 2 + 22 + 23 + 24 + ...... + 2100
=> 2A = 2 + 22 + 23 + 24 + ...... + 2101
=> 2A - A = 2101 - 1
=> A = 2101 - 1
\(a.2^x=128:4=32\)
\(\Rightarrow x=5\)
\(\text{b.(x - 6)3 = (x - 6)2}\)
\(\Rightarrow\hept{\begin{cases}x-6=1\\x-6=0\end{cases}\Rightarrow\hept{\begin{cases}x=7\\x=6\end{cases}}}\)
\(c.x=1;0\)
\(d.\text{ (7x - 11)3 = 25 . 52 + 200}\)
\(\text{ (7x - 11)3 = 1500}\)
\(\text{x là số thập phân}\)
\(\text{e. (2x - 2)3 = 8}\)
\(2x-2=2\)
\(x=2\)
\(\text{f, 3+2x-1 = 24 - [ 42 - ( 22 - 1) ]}\)
\(\text{f, 3+2x-1 =203 }\)
\(2x-1=200\)
\(2x=201\)
\(x=100.5\)
a) \(2^x\cdot4=128\)
\(2^{x+2}=2^7\)
\(x=5\)
b) \(\left(x-6\right)^3=\left(x-6\right)^2\)
\(\left(x-6\right)^3:\left(x-6\right)^2=1\)
\(x-6=1\)
\(x=7\)
c) \(x^{17}=x\)
\(x\in\left\{-1;0;1\right\}\)
d) \(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\left(7x-11\right)^3=1000\)
\(7x-11=10\)
\(7x=21\)
\(x=3\)
e) \(\left(2x-2\right)^3=8\)
\(2x-2=2\)
\(2x=4\)
\(x=2\)
f) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(3+2^{x-1}=11\)
\(2^{x-1}=8\)
\(x-1=3\)
\(x=4\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6