1. Tính giá trị
a,\(D=\left[\frac{1}{2^2}-1\right]\left[\frac{1}{3^2}-1\right]\left[\frac{1}{4^2}-1\right]...\left[\frac{1}{100^2}-1\right]\)
b, Tìm \(x\inℤ\)biết :
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}=\frac{220}{39}\)
c, Cho \(A=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\)
\(B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}\)
So sánh A và B
2.a,Cho M là trung điểm của đoạn thẳng AB.Trên tia đối của tia BA lấy điểm I sao cho \(IM=a(a>0)\)
Tính IA + IB theo a
b, Trên đường thẳng xy lấy điểm A vẽ hai tia Az và At,trên 2 nửa mặt phẳng đối nhau có bờ là xy sao cho góc \(\widehat{yAz}=\frac{1}{2}\widehat{yAt}\). Tìm số đo góc \(\widehat{yAz}\)để góc \(\widehat{xAz}=3\widehat{xAt}\)
\(D=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{99.101}{100^2}\)
\(=\frac{1.2...99}{2.3...100}.\frac{3.4....101}{2.3....100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
1 b) Đặt A=\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{66}+\frac{1}{78}\)
=> \(\frac{A}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}+\frac{1}{156}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{11.12}+\frac{1}{12.13}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}=\frac{1}{3}-\frac{1}{13}\)
=> \(A=\frac{2}{3}-\frac{2}{13}\)\(=\frac{20}{39}\)
Ta có: \(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+...+\frac{x}{78}=\frac{220}{39}\)
<=> \(x\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{15}+...+\frac{1}{78}\right)=\frac{220}{39}\Leftrightarrow x.\frac{20}{39}=\frac{220}{39}\Leftrightarrow x=11\)
\(B=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}\right)\)
\(< \left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)+\left(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\right)+\left(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\right)+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\left(\frac{1}{15}+\frac{1}{15}+\frac{1}{15}\right)\)\(=3.\frac{1}{3}+3.\frac{1}{6}+3.\frac{1}{9}+3..\frac{1}{12}+3.\frac{1}{15}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=3\)
=> B<3
\(A=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}=1-\frac{1}{2011}+1-\frac{1}{2012}+1+\frac{2}{2010}=3-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2010}+\frac{1}{2010}\)
\(>3-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2012}=3\)
=> A>3
Vậy A>3>B
Có thể chứng minh :
A>2,5>B
... nhiều cách khác nữa
2a)
Ta có: IA+IB=IB+AB+IB=AM+MB+2.IB=2.MB+2.IB=2(MB+IB)=2.MI=2.a
b)
Ta có: \(\widehat{yAt}+\widehat{tAx}=180^o\Rightarrow2.\widehat{yAz}+\frac{1}{3}.\widehat{zAx}=180^o\Rightarrow2.\widehat{yAx}+\frac{1}{3}.\left(180^o-\widehat{yAx}\right)=180^o\)
=> \(2.\widehat{yAx}+60^o-\frac{1}{3}.\widehat{yAx}=180^o\Rightarrow\frac{5}{3}\widehat{yAx}=120^o\Rightarrow\widehat{yAx}=72^o\)
bài 2b Bắt đầu từ dấu suy ra thứ 2 em sửa góc yAx thành yAz giúp cô nhé :)))
Chỗ : \(2\cdot\widehat{yAz}+\frac{1}{3}\left[180^o-\widehat{yAz}\right]\)hả cô
P/S : Em làm bài ngoài câu trả lời :v những câu đó dễ nên em sẽ làm luôn :v mong cô nhận xét
1. Tính :
\(\left[\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right]:\left[\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right]\)
Ta có : \(\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right]:\left[\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\underrightarrow{SC}\right]\)
Đặt biểu thức thứ 2 là số chia
\(SC=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left[1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]-2\left[\frac{1}{2}--\frac{1}{4}-\frac{1}{6}-....-\frac{1}{100}\right]\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=SBC\)
\(\Rightarrow SBC:SC=1\)
4.Chứng tỏ :
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
Ta có : \(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right]\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-2\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right]\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
