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2^1 + 2^2 + 2^3 +...+ 2^2010
= (2^1 + 2^2) + (2^3 + 2^4) + ... + (2^2009 + 2^2010)
= 2.(1 + 2) + 2^3.(1 + 2) + ... + 2^2009.(1 + 2) = 2.3 + 2^3.3 + ... + 2^2009.3 = 3.(2 + 2^3 + ... + 2^2009) => 2^1 + 2^2 + 2^3 +...+ 2^2010 chia hết cho 3 2^1 + 2^2 + 2^3 +...+ 2^2010 = (2^1 + 2^2 + 2^3) + ... + (2^2008 + 2^2009 + 2^2010) = 2.( 1 + 2 + 2^2) + ... + 2^2008.(1 + 2 + 2^2) = 2.7 + ... + 2^2008. 7 => 2^1 + 2^2 + 2^3 +...+ 2^2010 chia hết cho 7
\(A=1+2^2+2^3+...+2^{10}\)
\(\Leftrightarrow2A=2+2^3+2^4+...+2^{11}\)
\(\Leftrightarrow A=2^{11}-1\)
A=1+22+23+...+210A=1+22+23+...+210
⇔2A=2+23+24+...+211⇔2A=2+23+24+...+211
⇔A=211−1
\(P=sin^2x+3cos^2x=1-cos^2x+3cos^2x=1+2cos^2x=1+2.\left(\dfrac{1}{4}\right)^2=\dfrac{9}{8}\)
Ta có:
Tập hợp A:
\(A=\left\{1;3;5;7;9\right\}\)
Tập hợp B:
\(B=\left\{0;1;2;4;5;6;8\right\}\)
Mà: \(C=A\cup B\)
\(\Rightarrow C=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
⇒ Chọn D
1:
\(A=\left\{0;1;2;4\right\};B=\left\{0;2;3;5;7\right\}\)
\(A\cap B=\left\{0;2\right\}\)
\(A\cup B=\left\{0;1;2;4;3;5;7\right\}\)
\(A\text{B}=\left\{1;4\right\}\)
\(B\text{A}=\left\{3;5;7\right\}\)
2: \(A\text{B}=\left\{1;4\right\}\)
\(A=\left\{0;1;2;4\right\}\)
=>(A\B)\(\subset\)A
A\(A\B)
={0;1;2;4}\{1;4}
={0;2}
=\(A\cap B\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-7}{156}\)
\(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-17}{12}\)
\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-7}{55}\)
\(\dfrac{-34}{37}.\dfrac{74}{-85}=\dfrac{4}{5}\)
\(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{10}{7}\)
Chúc bạn học tốt!!!
a) \(\left(-\dfrac{1}{39}\right)+\left(-\dfrac{1}{52}\right)=\dfrac{-4-3}{156}=-\dfrac{7}{156}\)
b) \(\left(-\dfrac{6}{9}\right)+\left(-\dfrac{12}{16}\right)=-\dfrac{6}{9}-\dfrac{12}{16}=-\dfrac{17}{12}\)
c) \(-\dfrac{2}{5}-\left(-\dfrac{3}{11}\right)=-\dfrac{2}{5}+\dfrac{3}{11}=-\dfrac{7}{55}\)
d) \(\left(-\dfrac{34}{37}\right)\cdot\left(-\dfrac{74}{85}\right)=2\cdot\dfrac{2}{5}=\dfrac{4}{5}\)
e) \(\left(-\dfrac{5}{9}\right):\left(-\dfrac{7}{18}\right)=\dfrac{5}{9}\cdot\dfrac{18}{7}=5\cdot\dfrac{2}{7}=\dfrac{10}{7}\)
\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9\)
\(a=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)\)
\(a=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+2^7\left(1+2+2^2\right)\)
\(a=2.7+2^4.7+2^7.7=7\left(2+2^4+2^7\right)⋮7\left(đpcm\right)\)
A=(2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)
A=2(1+2+2^2)+2^4(1+2+2^2)+2^7(1+2+2^2)
A=2.7+2^4.7+2^7.7\(⋮\)7
Vậy A\(⋮\)7