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a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
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a) 998.34=(1000−2).34=34000−68=33932
b,\(1990.1990-\left(1990+2\right).1988=1990.1990-1990.1988-2.1988=1990\left(1990-1988\right)-2.1988=1990.2-2.1998=2\left(1990-1998\right)=2.2\)= 4
c. \(\text{(1374.57+678.86).(26.13+74.14)}\)
Câu này đề hơi lạ
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\(E\subset N\)
\(N=\left\{a;b\right\}=\left\{3;4\right\};\left\{3;5\right\};\left\{4;5\right\}\)
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1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
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Ư(4)= {-4;-2;-1;1;2;4}
Ư(-3)= {-3;-1;1;3}
Ư(12)={-12;-6;-4;-3;-2;-1;1;2;3;4;6;12}
Ư(-8)={-8;-4;-2;-1;1;2;4;8}
Ư(-6)={-6;-3;-2;-1;1;2;3;6}
Ư(-20)={-20;-10;-5;-4;-2;-1;1;2;4;5;10;20}
Ư(-10)={-10;-5;-2;-1;1;2;5;10}
Ư(-16)={-16;-8;-4;-2;-1;1;2;4;8;16}
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\(D=8+8^3+8^5+...+8^{2x+1}\)
\(\Rightarrow64D=8^3+8^5+...+8^{2x+3}\)
\(\Rightarrow63D=64D-D=8^3+8^5+...+8^{2x+3}-8-8^3-...-8^{2x+1}=8^{2x+3}-8\)
\(\Rightarrow8^{2x+3}-8+8=8^{51}\)
\(\Rightarrow8^{2x+3}=8^{51}\Rightarrow2x+3=51\Rightarrow2x=48\Rightarrow x=24\)
15⋮(2x+1)
=> 2x+1 ϵ Ư(15)
=> 2x+1 ϵ {1;3;5;15}
=> 2x ϵ {0;2;4;14}
=> x ϵ {0;1;2;7}
Vậy x ϵ {0;1;2;7}