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Bài 1a
B=4/1.3 + 4/3.5 + 4/5.7+...+4/2017.2019
B=4.2/(1.3).2 + 4.2/(3.5).2 + 4.2/(5.7).2+....+4.2/(2017.2019).2
B=2.( 2/1.3 + 2/3.5 + 2/5.7 +...+ 2/2017.2019 )
B=2.(1-1/3+1/3-1/5+1/5-1/7+....+1/2017-1/2019)
B=2.(1-1/2019)
B=2.(2019/2019-1/2019)
B=2.2018/2019
B=4036/2019
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
\(A=\frac{\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{100}}{1+2+3+...+2010}\)
\(\Leftrightarrow A=\frac{\left(-1\right)+1+\left(-1\right)+...+\left(-1\right)}{1+2+3+...+2010}\)
\(\Leftrightarrow A=\frac{0}{1+2+3+...+2010}\)
\(\Leftrightarrow A=0\)
Câu hỏi của Trần Thuỳ Trang - Toán lớp 6 - Học toán với OnlineMath
Bạn tham khảo.
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{39}}\)
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{40}}\)
=> 2A = 3A - A = \(1-\frac{1}{3^{40}}\)=> \(\frac{1-\frac{1}{3^{40}}}{2}=\frac{1}{2}-\frac{1}{3^{40}\cdot2}\)
Mấy câu còn là thì tương tự nhé c
câu b nhân vào \(2^2\)
câu c nhân vào 4
Đặt A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
3A=3/2.5+3/5.8+....+3/17.20
3A=1/2-1/5+1/5-1/8+...+1/17-1/20
3A=1/2-1/20
3A=9/20
2)
Giữ nguyên p/s 1/2^2
Ta có:1/3^2<1/2.3
1/4^2<1/3.4
...............
1/n^2<1/(n-1).n
=>1/3^2+1/4^2+...+1/n^2<1/2.3+1/3.4+...+1/(n-1).n
=>1/3^2+1/4^2+.....+1/n^2<1/2-1/3+1/3-1/4+.........+1/n-1-1/n
=>1/2^2+1/3^2+.....+1/n^2<1/2^2+1/2-1/n
=>1/2^2+1/3^2+....+1/n^2<3/4-1/n<3/4
3)
2B=2/3.5+2/5.7+....+2/47.49+2/49.51
2B=1/3-1/5+1/5-1/7+.....+1/47-1/49+1/49-1/51
2B=1/3-1/51
2B=16/51
B=16/51:2
B=8/51
A=1+1/2+1/2^2+...+1/2^2010
2A=2+1+1/2+....+1/2^2009
2A-A=(2+1+1/2+...+1/2^2009)-(1+1/2+1/2^2+....+1/2^2010)
A=2-1/2^2010