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a/ theo de bai ta co
4^24= (2^2)^24=2^48
8^17= (2^3)^17=2^51
vì 51>48 suy ra 2^51>2^48
cũng như 4^24<8^17
b/ theo đề bài ta có
3^300=(3^2)^150=9^150
2^450=(2^3)^150=8^150
vì 9>8 suy ra 9^150>8^150
cũng như 3^300>2^450
c/theo đề bài ta có
9^23=(3^2)^23=3^46
27^49=(3^3)^49=3^147
vì 147>46 suy ra 3^147>3^46
cũng như 9^23<27^49
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B
\(a,\frac{2}{3}>\frac{1}{4}\)
\(b,\frac{7}{10}< \frac{7}{8}\)
\(c,\frac{6}{7}>\frac{3}{5}\)
\(d,\frac{14}{21}< \frac{60}{72}\)
\(e,\frac{16}{9}< \frac{24}{13}\)
\(g,\frac{27}{82}< \frac{26}{75}\)
a 2/3 > 1/4
b 7/10 < 7/8
c6/7 > 3/5
d14/21 < 60/72
e16/9 < 24/13
g27/82<26/75
a/
\(9^5=\left(3^2\right)^5=3^{10}>3^9=\left(3^3\right)^3=27^3\)
b/ \(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}=\left(2^3\right)^{100}=2^{300}\)
c/
\(3.4^7=3.\left(2^2\right)^7=3.2^{14}>2.2^{14}=2^{15}=\left(2^3\right)^5=8^5\)
a) 2^27=(2^3)^9=8^9
3^18=(3^2)^9=9^9
Vì 8^9 bé hơn 9^9 nên 2^27 bé hơn 3^18
a: 27^4=(3^3)^4=3^12<3^18
b: 49^4=(7^2)^4=7^8
c: 9^16=(9^2)^8=81^8>27^8