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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
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Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a) \(2^x=32\)=>\(2^x=2^5\)=>\(x=5\)
b) \(64\times4^x=4^5\)
\(4^3.4^x=4^5\)
\(4^{3+x}=4^5\)
=>\(3+x=5\)
=> x=2
c)\(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
=>\(2^x=2^5\)
=>\(x=5\)
a, 2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
b, 64 x 4 mũ x = 4 mũ 5
64 x 4 mũ x = 1024
4 mũ x = 1024 : 64
4 mũ x = 16
4 mũ x = 4 mũ 2
x = 2
c, 2 mũ x - 15 = 17
2 mũ x = 17 + 15
2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5