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a) \(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}\)
b) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}\)
c) \(\frac{121.75.130.69}{39.60.11.198}=\frac{11^2.3.5^2.2.5.13.23.3}{13.3.2^3.3.5.11.11.3^2.2}\)
\(=\frac{11^2.3^2.5^3.2.13.23}{13.3^4.2^4.5.11^2}=\frac{5^2.23}{3^2.2^3}\)
a,\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2^2.2.3^2.3^2}{2^2.3^2.5}\)\(=\frac{2.3^2}{5}\)
b,\(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2^3.2.5^2.11.11.7}{2^3.5^2.5.7.7.11}=\frac{2.11}{5.7}\)
a) \(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{1.1.5}=\frac{18}{5}\)
b) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.1.11.1}{1.5.7.1}=\frac{22}{35}\)
Bài 1:
a) Ta có: 536=(53)12=12512
1124=(112)12=12112
Vì 12512>12112
=>536>1124
b) Ta có: 6255=(54)5=520
1257=(53)7=521
Vì 520<521
=>6255<1257
c) Ta có: 32n=(32)n=9n
23n=(23)n=8n
Vì 9n>8n
=>32n>23n
d) Ta có: 6.522=(1+5).522=523+522>523
e) S=1+2+22+23+...+22005
2S=2+22+23+24+...+22006
=>2S-S=(2+22+23+24+...+22006) - (1+2+22+23+...+22005)
=>S=22006-1<22014<5.22014
Cậu cho tớ 3 tớ sẽ làm 2 bài còn lại cho cậu
a/ \(\dfrac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\dfrac{2\cdot3^2}{5}=\dfrac{18}{5}\);
\(\dfrac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\dfrac{2\cdot11}{5\cdot7}=\dfrac{22}{35}\)
b/ \(\dfrac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\dfrac{11^2\cdot\left(3\cdot5^2\right)\cdot\left(2\cdot5\cdot13\right)\cdot13^2}{\left(3\cdot13\right)\cdot\left(2^2\cdot3\cdot5\right)\cdot11\cdot\left(2\cdot3^2\cdot11\right)}\)
\(=\dfrac{11^2\cdot2\cdot3\cdot5^3\cdot13^3}{11^2\cdot2^3\cdot3^4\cdot5\cdot13}=\dfrac{5^2\cdot13^2}{2^2\cdot3^3}=\dfrac{4225}{108}\)
c/ \(\dfrac{1989\cdot1990+3978}{1992\cdot1991-3984}=\dfrac{1989\cdot1990+2\cdot1989}{1992\cdot1991-2\cdot1992}\) (sửa đề)
\(=\dfrac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991-2\right)}=\dfrac{1989\cdot1992}{1992\cdot1989}=1\)
d/ \(\dfrac{125}{1000}=\dfrac{125:125}{1000:125}=\dfrac{1}{8}\);
\(\dfrac{198}{126}=\dfrac{198:18}{126:18}=\dfrac{11}{7}\);
\(\dfrac{3}{243}=\dfrac{3:3}{243:3}=\dfrac{1}{81}\);
\(\dfrac{103}{3090}=\dfrac{103:103}{3090:103}=\dfrac{1}{30}\)