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a) \(B=1+3+3^2+3^3+....+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+2^3\right)+....+3^{96}\left(1+3+3^2+3^3\right)\)
\(=\left(1+3+3^2+3^3\right)\left(1+3^4+...+3^{96}\right)\)
\(=40\left(1+3^4+....+3^{96}\right)\)\(⋮\)\(40\)
b) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=40.3^4\)
bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
Bài 1 :
a) \(a.b+b.19=713\) \(\left(a;b\inℕ^∗\right)\)
\(\Rightarrow b.\left(a+19\right)=713\)
\(\Rightarrow\left(a+19\right);b\in\left\{1;23;31;713\right\}\)
\(\Rightarrow\left(a;b\right)\in\left\{\left(-18;713\right);\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\)
\(\Rightarrow\left(a;b\right)\in\left\{\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\left(a;b\inℕ^∗\right)\)
b) \(a.b-10.b=650\)
\(\Rightarrow b.\left(a-10\right)=650\)
\(\Rightarrow\left(a-10\right);b\in\left\{1;5;10;13;25;26;50;65;130;325;650\right\}\)
Bạn lập bảng sẽ tìm ra (a;b)...
Bài 2 :
a) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=3^4.40\)
b) \(B=1+3+3^2+3^3+...+3^{99}\)
\(\Rightarrow B=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)...+3^{96}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow B=40+3^4.40...+3^{96}.40\)
\(\Rightarrow B=40\left(1+3^4...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)