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Xét hiệu:
\(C=1-\dfrac{7^{2011}+1}{7^{2013}+1}=\dfrac{7^{2011}\left(7^2-1\right)}{7^{2013}+1}=\dfrac{48.7^{2011}}{7^{2013}+1}\)
\(D=1-\dfrac{7^{2013}+1}{7^{2015}+1}=\dfrac{7^{2013}\left(7^2-1\right)}{7^{2015}+1}=\dfrac{48.7^{2013}}{7^{2015}+1}\)
Ta có:
\(\dfrac{C}{D}=\dfrac{48.7^{2011}}{7^{2013}+1}\cdot\dfrac{7^{2015}+1}{48.7^{2013}}=\dfrac{7^{2015}+1}{\left(7^{2013}+1\right)\cdot7^2}=\dfrac{7^{2015}+1}{7^{2015}+49}< 1\)
=> C<D =>A>B
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
Câu 2;3;4 dễ quá... bỏ qua!!
Câu 5;6 khó quá ... khỏi làm!!
dễ quá bỏ qua!!, khó quá khỏi làm!!
cứ tiêu chí mày bạn sẽ vượt qua mọi bài toán... và nhanh chóng đạt 1đ.
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
a: Số số hạng là 2014-1+1=2014 số
A=2014*2015/2=2029105
b: Số số hạng là (2013-1):2+1=1007(số)
B=(2013+1)*1007/2=1014049
c: Số số hạng là (2014-2):2+1=1007(số)
Tổng là (2014+2)*1007/2=1015056
d: Số số hạng là (2014-1):3+1=672(số)
Tổng là (2014+1)*672/2=677040
e: Số số hạng là (2015-5):5+1=403(số)
Tổng là (2015+5)*403/2=407030
b, Ta có:
\(14A=\dfrac{7^{2013}+14}{7^{2013}+1}=\dfrac{7^{2013}+1+13}{7^{2013}+1}=\dfrac{7^{2013}+1}{7^{2013}+1}+\dfrac{13}{7^{2013}+1}=1+\dfrac{13}{7^{2013}+1}\)
\(14B=\dfrac{7^{2015}+14}{7^{2015}+1}=\dfrac{7^{2015}+1+13}{7^{2015}+1}=\dfrac{7^{2015}+1}{7^{2015}+1}+\dfrac{13}{7^{2015}+1}=1+\dfrac{13}{7^{2015}+1}\)
\(\)Vì \(7^{2013}+1< 7^{2015}+1\)
\(\dfrac{\Rightarrow13}{7^{2013}+1}>\dfrac{13}{7^{2015}+1}\)
\(\Rightarrow1+\dfrac{13}{7^{2013}+1}>1+\dfrac{13}{7^{2015+1}}\)
\(\Leftrightarrow14A>14B\)
\(\Rightarrow A>B\)