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=>B=\(\dfrac{1}{4.4}+\dfrac{1}{6.6}+\dfrac{1}{8.8}+...+\dfrac{1}{2006.2006}\)
=>B<\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2005.2007}\)
=>B<\(\dfrac{2}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2005.2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2005.2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{2005}-\dfrac{1}{2005}-\dfrac{1}{200}\right)\)(xin lỗi, đoạn cuối (chỗ 200 í )là 2007 nhá
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{2007}\right)\)
=>B<\(\dfrac{1}{2}.\dfrac{668}{2007}\)
=>B<\(\dfrac{1.668}{2.2007}\)
=>B<\(\dfrac{1.668:2}{2.2007:2}\)
=>B<\(\dfrac{334}{2007}\)
Tick cho tôi nha :D
Ta thấy : \(\frac{1}{4^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{2006^2}< \frac{1}{2005.2006}\)
\(\Rightarrow B=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2006^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
\(\Leftrightarrow B< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Leftrightarrow B< \frac{1}{4}-\frac{1}{2006}=\frac{1001}{4012}\)
Mà \(\frac{1001}{4012}< \frac{334}{2007}\Rightarrow B< \frac{334}{2007}\)
\(B< \frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2006.2008}\)
\(2B< \frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2006.2008}=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2008}=\frac{1}{4}-\frac{1}{2008}=\frac{501}{2008}\)\(B< \frac{501}{4016}< \frac{501}{4014}< \frac{668}{4014}=\frac{334}{2007}\)
Vậy:.....
Mai ơi! bạn khùng hả? ko trả lời thì thôi lại còn vào chỗ trả lời để sorry
A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)
A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)
Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)
nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)
\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)
A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)
A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)
\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\)
Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A> \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)
Tương tự ta có :
A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)
A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)
A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)
\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)
2x(3y-2)+(3y-2) = (2x+1)(3y-2) = -55.Lập bảng :
2x+1 | -55 | -11 | -5 | -1 | 1 | 5 | 11 | 55 |
3y-2 | 1 | 5 | 11 | 55 | -55 | -11 | -5 | -1 |
2x | -56 | -12 | -6 | -2 | 0 | 4 | 10 | 54 |
3y | 3 | 7 | 13 | 57 | -53 | -9 | -3 | 1 |
x | -28 | -6 | -3 | -1 | 0 | 2 | 5 | 27 |
y | 1 | 19 | -3 | -1 |
Vậy (x;y) = (-28;1);(-1;19);(2;-3);(5;-1)
Ta có : \(\frac{334}{2007}\)nhỏ hơn 1 vì 334 < 2007
Mà mọi số hạng trong tổng 142 + 162 + 182 + .... + 120062 chắc chắn sẽ lớn hơn 1 => Vô lí
=> Đề có thể sai
a)[98+(-100)+102].(-2)
=(98-100+102).(-2)
=98-100+102.(-2)
=98-100+(-204)
=-2+(-204)
=-206
b)245+[37+(-245)-|-17|]
=245+(37-245-17)
=245+37-245-17
=245-245+37-17
=0+37-17
=37-17
=20
c)|5|-(-5)-|-7|.3
=5+5-7.3
=5+5-21
=10-21
=11
d)-(2-5)-13
=-2+5-13
=3-13
=-10
e)334+(-20)+(2007-334)
=334-20+2007-334
=334-20-(-2007)-334
=334-334+2007-20
=0+2007-20
=2007-20
=1987
f)-1+3-5+7-9+...+103
=-1+3+(-5)+7+(-9)+...+103
=-2+(-2)+(-2)+...+-2(Có tất cả 51 số -2)
=-2.51
=-102
Xog r` đó