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a) Ta có : \(31^5< 32^5=\left(2^5\right)^5=2^{25}< 2^{28}=\left(2^4\right)^7=16^7< 17^7\)
\(\Rightarrow31^5< 17^7\)
b) Ta có : \(8^{12}=\left(2^3\right)^{12}=2^{36}>2^{32}=\left(2^4\right)^8=16^8>12^8\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{99}\)
\(A=\frac{1-\frac{1}{99}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
a) \(31^5< 34^5=2^5.17^5=32.17^5\)
\(17^7=17^2.17^5=289.17^5\)
\(\Rightarrow31^5< 17^7\)
b) \(12^8< 16^8=\left(2^4\right)^8=2^{32}\)
\(8^{12}=\left(2^3\right)^{12}=2^{36}\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{1}{3^{99}}\)
\(\Rightarrow2A=1-\frac{1}{3^{99}}< 1\Rightarrow A< \frac{1}{2}\)
P = 1 + 32 + 34 + 36+......+3100
32 P= 32(1 + 32 + 34 + 36+......+3100)
32P= 32 + 34 + 36+......+3100+3102
32P= (32 + 34 + 36+......+3100+3102)- (1 + 32 + 34 + 36+......+3100 )
32 P= 3102 - 1
P= (3102 -1) :9
Q = (917)3 / 23
Q = 951 / 8
Q = (32)51 /8
Q = 3102 /8
Q= 3102 :8
=> P > Q
Vậy...
K chắc nha b
xét P=1+3^2+3^4+3^6+3^8+....+3^100
=> 3^2.P=3^2+3^4+3^6+3^8+3^10+...+3^102
9.P-P=(3^2+3^4+3^6+3^8+3^10+...+3^102)-(1+3^2+3^4+3^6+3^8+....+3^100)
8P=3^102-1
P=\(\frac{3^{102}-1}{8}\)
Xét Q :
\(\left(\frac{9^{17}}{2}\right)^3=\left[\frac{\left(3^2\right)^{17}}{2}\right]^3=\frac{\left(3^{34}\right)^3}{8}=\frac{3^{102}}{8}\)
mà 3^102-1<3^102
=>P<Q