1+4+7+10+...+243
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1: \(=\dfrac{272-168+186}{30}\cdot\dfrac{7}{9}=\dfrac{29}{3}\cdot\dfrac{7}{9}=\dfrac{203}{27}\)
2: \(=\dfrac{9-55}{33}\cdot\dfrac{6-5}{8}-\dfrac{4}{5}\cdot\dfrac{1}{24}=\dfrac{-4}{3}\cdot\dfrac{1}{8}-\dfrac{1}{30}=\dfrac{-1}{6}-\dfrac{1}{30}=\dfrac{-6}{30}=-\dfrac{1}{5}\)
3: \(=\dfrac{7}{3}\left(\dfrac{6}{25}+\dfrac{19}{25}\right)=\dfrac{7}{3}\cdot1=\dfrac{7}{3}\)
\(\left(\frac{1}{27}\right)^{10}\&\left(\frac{1}{243}\right)^7\)
\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)
\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)
Vậy \(\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)
Ta có :
\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)
\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)
Do : \(\frac{1}{3^{30}}>\frac{1}{3^{35}}\left(3^{30}< 3^{35}\right)\)
\(\Rightarrow\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)
=x^2.x^8.x^4.x^6.x^10/x^1.x^9.x^3.x^7.x^3=3^5
=x^10.x^10.x^10 / x^10.x^10.x^5=3^5
=x^10/x^5=3^5
=x^5=3^5
x=3
\(x^2.x^4.x^6.x^8.x^{10}:\left(x.x^3.x^5.x^7.x^9\right)=243\)
\(x^{2+4+6+8+10}:x^{1+3+5+7+9}=243\)
\(x^{30}:x^{25}=243\)
\(x^{30-25}=243\)
\(x^5=243\)
\(x^5=3^5\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
1: 243^5=(3^5)^5=3^25
3*27^8=3*(3^3)^8=3^25
=>243^5=3*27^8
6: 125^5=(5^3)^5=5^15
25^7=(5^2)^7=5^14
=>125^5>25^7(15>14)
5: 78^12-78^11=78^11(78-1)=78^11*77
78^11-78^10=78^10*77
mà 11>10
nên 78^12-78^11>78^11-78^10
1: 8=2^3
2: 25=5^2
3: 4=2^2
4: 49=7^2
5: 81=9^2
6: 36=6^2
7: 100=10^2
8: 121=11^2
9: 144=12^2
10: 169=13^2
11: 27=3^3
12: 125=5^3
13: 1000=10^3
14: 32=2^5
15: 243=3^5
16: 343=7^3
17: 216=6^3
18: 64=4^3
19: 225=15^2
20: 128=2^7
số số hạng là:243 :3=81 số
tổng dãy số là:{ 243+1}x81:2=9882
đ/s:9882
Bạn ơi, đề bài sai rồi!