Bài 1:
Cách 1:
\(A=\left\{0;2;4;6;8;10;12\right\}\)
Cách 2:
\(A=\left\{x\in N|x=2k;0\le k\le6;k\in N\right\}\)
Bài 2:
a) \(3^x\cdot3=243\)
\(\Rightarrow3^{x+1}=3^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=5-1\)
\(\Rightarrow x=4\)
b) \(2^{x+2}-2^x=96\)
\(\Rightarrow2^x\cdot\left(2^2-1\right)=96\)
\(\Rightarrow2^x\cdot3=96\)
\(\Rightarrow2^x=\dfrac{96}{3}\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Bài `3`
\(a,27\cdot35+27\cdot65-170\\ =27\cdot\left(35+65\right)-170\\ =27\cdot100-170\\ =2700-170\\ =2530\\ b,142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\\ =142-\left[50-\left(8\cdot10-8\cdot5\right)\right]\\ =142-\left[50-\left(80-40\right)\right]\\ =142-\left(50-40\right)\\ =142-10\\ =132\)
Bài 3:
a) \(27\cdot35+27\cdot65-170\)
\(=27\cdot\left(35+65\right)-170\)
\(=27\cdot100-170\)
\(\Rightarrow2700-170\)
\(=2530\)
b) \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)
\(=142-\left[50-\left(8\cdot10-8\cdot5\right)\right]\)
\(=142-\left[50-\left(80-40\right)\right]\)
\(=142-\left(50-40\right)\)
\(=142-10\)
\(=132\)
