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\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)(1)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}\)
Thay A vào (1)
\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)
\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Ta có: 210 < 211
\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
Bài 2:
a: \(3B=3+3^2+3^3+...+3^{90}\)
\(\Leftrightarrow2B=3^{90}-1\)
hay \(B=\dfrac{3^{90}-1}{2}\)
b: \(B=\left(1+3+3^2+3^3+3^4+3^5\right)+3^6\left(1+3+3^2+3^3+3^4+3^5\right)+...+3^{84}\left(1+3+3^2+3^3+3^4+3^5\right)\)
\(=384\cdot\left(1+3^6+...+3^{84}\right)⋮52\)
a: \(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
b: \(=3^4-\left(-8\right)^2-\left(-25\right)^2\)
\(=81-64-625=-608\)
c: \(=2^3+3\cdot1\cdot\dfrac{1}{4}\cdot4+\left[4:\dfrac{1}{2}\right]:8\)
\(=8+3+4\cdot2:8=11+1=12\)
\(a,7^6+7^5-7^4=7^4\left(7^2+7-1\right)\\ =7^4\cdot55\\ \Rightarrow7^6+7^5-7^4⋮55\)
\(b,3^{n+2}-2^{n+2}+3^n-2^n\\ =3^n\cdot3^2+3^n-2^n\cdot2^2-2^n\\ =3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\\ =3^n\cdot10-2^{n-1}\cdot2\cdot5\\ =10\cdot\left(3^n-2^{n-1}\right)\\ \Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(c,8^7-2^{18}=8^7-\left(2^3\right)^6\\ =8^7-8^6\\ =8^6\cdot\left(8-1\right)\\ =8^5\cdot8\cdot7\\ =8^5\cdot4\cdot14\\ \Rightarrow8^7-2^{18}⋮14\)
b,\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n.\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(\Rightarrow D=1-\frac{1}{n+2}=\frac{n}{n+2}< \frac{n+2}{n+2}=1\left(1\right)\)
\(\Rightarrow D=\frac{n}{n+2}>0\left(2\right)\)
Từ (1);(2)\(\Rightarrow0< D< 1\)
\(\Rightarrowđpcm\)
a,\(C>0\)
\(C=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}< 9;\frac{1}{11}< 1\)
\(\Rightarrow0< A< 1\)
\(\Rightarrow A\notinℤ\)
c,\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Ta quy đồng 3 số đầu
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>\frac{6.2}{12}=1\)
\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< \frac{6.2}{6}=2\)
\(1< E< 2\)
\(E\notinℤ\)
Đặt:
\(A=\left(2^{10}+2^{11}+2^{12}\right)\)
\(A=\left(2^{10}+2^{10}.2+2^{10}.2^2\right)\)
\(A=2^{10}\left(2+2^2+1\right)\)
\(A=2^{10}.7\)
\(\Leftrightarrow A⋮7\)
\(2^{10}+2^{11}+2^{12}=2^{10}\left(1+2+2^2\right)=2^{10}.7⋮7\)
\(\Rightarrowđpcm\)