Tính theo cách hợp lí:
A=3+6+9+12+...+2007
B=\(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
C= \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+.....+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+.....+\dfrac{1}{2006}}\)
GIÚP mình nha
Lèm ơn đấy !!!!!
Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)
\(=\dfrac{2006}{2007}\)
TÍNH NHANH:
C= \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4} +....+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+....+\dfrac{1}{2006}}\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
a,tính tổng : \(S=\dfrac{27+4500+135+550+2}{2+4+6+...+14+16+18}\)
b, So sánh : \(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}v\text{à }B=\dfrac{2006^{2005}+1}{2006^{2006}+1}\)
- Mình dùng cách lớp 8 để làm câu b được không :)?
- Tham khảo câu b:
https://olm.vn/hoi-dap/tim-kiem?q=+++++++++++A=2006%5E2005+1/2006%5E2006+1B=2006%5E2006+1/2006%5E2007+1so+s%C3%A1nh+A+v%C3%A0+B&id=520258
Rút gọn:
a) \(A=\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+... +\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2006\sqrt{2005}+2005\sqrt{2006}}+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\)
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
Tiếp phần b ( do máy lag) :3
Cộng 2 vế với nhau, ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\\ =1-\dfrac{1}{\sqrt{2007}}\)
a) A=\(\dfrac{1}{\sqrt{3}+\sqrt{5}}\)+\(\dfrac{1}{\sqrt{5}+\sqrt{7}}\)+\(\dfrac{1}{\sqrt{7}+\sqrt{9}}\)+...+\(\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)+\(\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)+\(\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\left(\sqrt{9}-\sqrt{7}\right)}\)+...+\(\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}}{2}\)
=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
vậy A=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}\)
B = \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
Tính \(\dfrac{B}{A}\)
Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)
\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)
\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)
Bài 1: tính M :N biết:
a) N=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.......+\dfrac{1}{2007}+\dfrac{1}{2008}\) và M= \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+.......+\dfrac{2}{2006}+\dfrac{1}{2007}\)
b) M= \(\dfrac{1}{11.13}+\dfrac{1}{13.15}+\dfrac{1}{15.17}+.......+\dfrac{1}{31.33}+\dfrac{1}{33.35}\) và N= \(\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+\dfrac{12}{15.17.19}+.....+\dfrac{12}{31.33.35}\)
a, Theo bài ra ta có:
\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)
( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'
\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)
\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)
Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:
\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(=>M:N=2008\)
Câu b đợi 1 chút nha.......
b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{33}\)
\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)
\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)
\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)
\(=\dfrac{92}{5005}\)
\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)
Vậy...
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\)
Tính
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+4=0\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+1+1+1+1=0\\ \Rightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{2006}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2005}+\dfrac{1}{2006}\right)=0\)
mà `1/2008+1/2007+1/2005+1/2006≠ 0`
`=> x+2010=0`
`=>x=-2010`
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\)
=>x+2010=0
=>x=-2010
Bài 2: Tính hợp lý:
\(A=\dfrac{63636337-37373763}{1+2+3+...+2006}\)
\(B=1\dfrac{6}{41}\left(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\right)\dfrac{124242423}{237373735}\)
\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)
\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)
=0
BT7: So sánh
2) \(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}\) và \(B=\dfrac{2006^{2005}+1}{2006^{2006}+1}\)
Ta có:
\(2006A=\dfrac{2006^{2007}+2016}{2006^{2007}+1}=1+\dfrac{2005}{2006^{2007}+1}\)
\(2006B=\dfrac{2006^{2006}+2006}{2006^{2006}+1}=1+\dfrac{2005}{2006^{2006}+1}\)
Do \(\dfrac{2005}{2006^{2006}+1}>\dfrac{2005}{2006^{2007}+1}\Rightarrow1+\dfrac{2005}{2006^{2006}+1}>1+\dfrac{2005}{2006^{2007}+1}\)
\(\Rightarrow2006A< 2006B\Rightarrow A< B\)
Mình sẽ giải cách ngắn hơn cách bạn đạt nha:
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}< 1\)
\(A< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}\Rightarrow A< \dfrac{2006^{2006}+2006}{2006^{2007}+2006}\Rightarrow A< \dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}\Rightarrow A< \dfrac{2006^{2005}+1}{2006^{2006}+1}=B\)\(A< B\)