Câu 1 So sánh
B = 3 75
C = 2100
Câu 2 Tìm x biết
{ [ ( 50 - x ) : 2 + 80 ] - 52 . 2 } . 4 = 200
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a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)}{2.2}=\frac{3\left(y-2\right)}{3.3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}=\frac{\left(2x+3y-z\right)-5}{9}=\frac{50-5}{9}=5\)
\(\Rightarrow x-1=10;y-2=15;z-3=20\)
\(\Rightarrow x=11;y=17;z=23\)
\(a,2^{x+1}=32\\ 2^{x+1}=2^5\\ x+1=5\\ x=4\\ b,2^{2x}+2^{2x+1}=48\\ 2^{2x}+2\cdot2^{2x}=48\\ 3\cdot2^{2x}=48\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=2\)
\(c,3^x+5\cdot3^{x+1}=144\\ 3^x+15\cdot3^x=144\\ 16\cdot3^x=144\\ 3^x=9\\ 3^x=3^2\\ x=2\\ d,3^{x+5}=9^{x+1}\\ 3^{x+5}=3^{2x+2}\\ x+5=2x+2\\ x=3\)
Câu 1: Ta có: \(B=3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(C=2^{100}=\left(2^4\right)^{25}=16^{25}\)
Vì 27 > 16 nên 2725>1625
Vậy B > C
Câu 2: \(\left\{\left[\left(50-x\right):2+80\right]-5^2.2\right\}.4=200\)
\(\Leftrightarrow\left[\left(50-x\right):2+80-50\right].4=200\)
\(\Leftrightarrow\left(50-x\right):2+30=50\)
\(\Leftrightarrow\left(50-x\right):2=20\)
\(\Leftrightarrow50-x=40\)
\(\Leftrightarrow x=10\)
Vậy x = 10
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(2^{100}=\left(2^4\right)^{25}=8^{25}\)
Vì \(27^{25}>8^{25}\)
\(\Rightarrow3^{75}>2^{100}\)