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Mk làm bai 1 thôi:
\(A=1+2+2^2+2^3+...+2^{2015}+2^{2016}\)
\(2A=2+2^2+2^3+2^4+...+2^{2016}+2^{2017}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}+2^{2017}\right)-\left(1+2+2^2+2^3+2^4+...+2^{2015}+2^{2016}\right)\)
\(A=2+2^2+2^3+2^4+...+2^{2016}+2^{2017}-1-2-2^2-2^3-2^4-...-2^{2016}-2^{2017}\)
\(A=2^{2017}-1\)
1/a,
-Ta có:
$B<1\Leftrightarrow B<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10(10^{2004}+1)}{10(10^{2005}+1)}=\frac{10^{2004}+1}{10^{2005}+1}=A$
-Vậy: B<A
b,$A=1+(\frac{1}{2})^2+...+(\frac{1}{100})^2$
$\Leftrightarrow A=1+\frac{1}{2^2}+...+\frac{1}{100^2}$
$\Leftrightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$
$\Leftrightarrow A<1+\frac{1}{1}-\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}$
$\Leftrightarrow A<1+1-\frac{1}{100}\Leftrightarrow A<2-\frac{1}{100}\Leftrightarrow A<2(đpcm)$
2,
a.
-Ta có:$\Rightarrow \frac{3x+7}{x-1}=\frac{3(x-1)+16}{x-1}=\frac{3(x-1)}{x-1}+\frac{16}{x-1}=3+\frac{16}{x-1}
-Để: 3x+7/x-1 nguyên
-Thì: $\frac{16}{x-1}$ nguyên
$\Rightarrow 16\vdots x-1\Leftrightarrow x-1\in Ư(16)\Leftrightarrow ....$
b, -Ta có:
$\frac{n-2}{n+5}=\frac{n+5-7}{n+5}=1-\frac{7}{n+5}$
-Để: n-2/n+5 nguyên
-Thì: \frac{7}{n+5} nguyên
$\Leftrightarrow 7\vdots n+5\Leftrightarrow n+5\in Ư(7)\Leftrightarrow ...$
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\frac{4949}{9900}\)
\(A=\frac{4949}{19800}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
A=\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+....+\frac{5^2}{26.31}\)
=>A=5.(\(\frac{5}{1.6}+\frac{5}{6.11}+....+\frac{5}{26.31}\))
=>A=5.(\(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\))
=>A=5.(\(\frac{1}{1}-\frac{1}{31}\))
=>A=5.\(\frac{30}{31}\)
=>A=\(\frac{150}{31}\)
=>A>1( vì tử của A lớn hơn mẫu )
a, gọi ƯCLN(14n+3;21n+5)=d
=> \(\left\{{}\begin{matrix}14n+3\\21n+5\end{matrix}\right.\)⋮d =>\(\left\{{}\begin{matrix}3\left(14n+3\right)\\2\left(21n+5\right)\end{matrix}\right.\)⋮d=>\(\left\{{}\begin{matrix}42n+9\\42n+10\end{matrix}\right.\)⋮d
=>(42n+10)-(42n+9)⋮d
=>1⋮d
=>d=1
Do ƯCLN của 14n+3 ; 21n+5 là 1
=> 2 số trên là hai số nguyên tố cùng nhau
=>hai số đó nếu chia cho nhau thì sẽ ko chia hết
=> hai số đó khi biểu diễn ở dạng phân số thì sẽ thành phân số tối giản
2.
Ta có : \(A=\frac{n+5}{n+2}=\frac{n+2+3}{n+2}=1+\frac{3}{n+2}\)
để A là số nguyên thì \(\frac{3}{n+2}\)là số nguyên
\(\Rightarrow3⋮n+2\)
\(\Rightarrow\)n + 2 \(\in\)Ư ( 3 ) = { 1 ; -1 ; 3 ; -3 }
Lập bảng ta có :
Vậy n \(\in\){ -1 ; -3 ; 1 ; -5 }
3.
\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
\(=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)
\(=97+\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
gọi \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)( 1 )
\(3B=1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)( 2 )
Lấy ( 2 ) trừ ( 1 ) ta được :
\(2B=1-\frac{1}{3^{98}}< 1\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{98}}}{2}< \frac{1}{2}< 1\)
\(\Rightarrow97+\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)< 100\)
4.
đặt \(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(5A=1-\frac{1}{31}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{31}}{5}< \frac{1}{5}< 1\)
Ta có : \(2A=2.\left(1+2+2^2+2^3+...+2^{2015}+2^{2016}\right)\)
\(2A=2+2^2+2^3+2^4+...+2^{2016}+2^{2017}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}+2^{2017}\right)-\left(1+2+2^2+2^3+...+2^{2015}+2^{2016}\right)\)
\(A=2+2^3+2^4+2^5+...+2^{2016}+2^{2017}-1-2-2^2-2^3-...-2^{2015}-2^{2016}\)
\(A=2^{2017}-1\)