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\(B=\left(1+\frac{1}{1.3}\right)+\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...+\left(1+\frac{1}{98.100}\right)\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\right)\)( 98 số 1 ở tồng đầu tiên)
\(=98+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.101}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=98+\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{3}{97.101}\right)+\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=98+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{98}-\frac{1}{100}\right)\)\(=98+\frac{1}{2}.\left(1-\frac{1}{101}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=98+\frac{1}{2}.\frac{100}{101}+\frac{1}{2}.\frac{49}{100}\)
\(=98+\frac{51}{101}+\frac{49}{200}\)
Suy ra phàn nguyên của B là 98.
Vậy phân fnguyên của B là 98.
mình nhầm. bạn thay các chỗ có "97.101" thành "99.101" nhé!
Xét : \(\frac{x^2}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{x^2-1}\)
=> \(\left[\frac{x^2}{x^2-1}\right]=1\) vì \(0< \frac{1}{x^2-1}< 1\)
Do đó : \(\left[D\right]=1.98=98\)
=>\(T=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{98^2}{97.99}.\frac{99^2}{98.100}\)
=>\(T=\frac{2^2.3^2.4^2...98^2.99^2}{1.3.2.4.3.5...97.99.98.100}\)
Trông thì khó vậy nhưng thực ra ko khó đâu, bạn chỉ việc rút gọn từ trên tử xuống dưới mẫu là xong
=>\(T=\frac{2.99}{1.100}=\frac{99}{50}=1\frac{49}{50}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{3.5}....\frac{98.98}{97.99}.\frac{99.99}{98.100}\)
\(=\frac{2.3.4....98.99}{1.3.5...97.98}.\frac{2.3.4....98.99}{3.5.7...99.100}\)
rút gọn đi có :
\(\frac{99}{1}.\frac{2}{100}=99.\frac{1}{50}=\frac{99}{50}\)
\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A
\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)
\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)
Dễ thấy phần nguyên của B là 98
Vậy phần nguyên của B là 98
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
*\(\frac{x}{200}\)=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\)....\(\frac{99^2}{99.100}\)
=>\(\frac{x}{200}\)=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{99}{100}\)
=>\(\frac{x}{200}\)=\(\frac{1}{100}\)
=>100x=200
=>x=2
Có sử dụng 1 hằng đẳng thức lớp 8 nhé : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+...+\frac{99^2}{98.100}\)
\(B=\frac{2^2}{\left(2-1\right)\left(2+1\right)}+\frac{3^2}{\left(3-1\right)\left(3+1\right)}+...+\frac{99^2}{\left(99-1\right)\left(99+1\right)}\)
\(B=\frac{2^2}{2^2-1}+\frac{3^2}{3^2-1}+...+\frac{99^2}{99^2-1}\)
\(B=\frac{2^2-1}{2^2-1}+\frac{1}{2^2-1}+\frac{3^2-1}{3^2-1}+\frac{1}{3^2-1}+...+\frac{99^2-1}{99^2-1}+\frac{1}{99^2-1}\)
\(B=1+\frac{1}{1.3}+1+\frac{1}{2.4}+...+1+\frac{1}{98.100}\)
\(B=\left(1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{98.100}\right)\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{98.100}\) ta có :
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{98.100}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}\left[\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{98}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\)
\(A=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}+\frac{1}{4}-\frac{1}{198}-\frac{1}{200}\)
\(\Rightarrow\)\(B=98+A=98+\frac{1}{2}+\frac{1}{4}-\frac{1}{198}-\frac{1}{200}=98+\left[\left(\frac{1}{2}+\frac{1}{4}\right)-\left(\frac{1}{198}+\frac{1}{200}\right)\right]\)
Ta có :
\(\frac{1}{2}>\frac{1}{198}\)
\(\frac{1}{4}>\frac{1}{200}\)
\(\Rightarrow\)\(\frac{1}{2}+\frac{1}{4}>\frac{1}{198}+\frac{1}{200}\)
\(\Rightarrow\)\(\left(\frac{1}{2}+\frac{1}{4}\right)-\left(\frac{1}{198}+\frac{1}{200}\right)>0\) \(\left(1\right)\)
Lại có :
\(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}< 1\)
\(\Rightarrow\)\(\left(\frac{1}{2}+\frac{1}{4}\right)-\left(\frac{1}{198}+\frac{1}{200}\right)< 1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(0< \left(\frac{1}{2}+\frac{1}{4}\right)-\left(\frac{1}{198}+\frac{1}{200}\right)< 1\)
\(\Rightarrow\)\(B=98+\left[\left(\frac{1}{2}+\frac{1}{4}\right)-\left(\frac{1}{198}+\frac{1}{200}\right)\right]\) có phần nguyên là \(98\)
Vậy \(B\) có phần nguyên là \(98\)
Chúc bạn học tốt ~
bạn giải theo cách lớp 6 đc ko vì mk mới học lớp 6 thôi