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\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=-\frac{98}{100}=-\frac{49}{50}\)
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=\frac{1}{100}-\left(\frac{1}{100}-1\right)\)
\(=1\)
Bài làm:
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(=\frac{1}{99.100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{98-97}{97.98}+\frac{99-98}{98.99}\right)\)
\(=\frac{1}{99.100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99.100}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99.100}-\frac{98}{99}\)
\(=\frac{1-98.100}{99.100}=\frac{1-9800}{9900}=-\frac{9799}{9900}\)
Học tốt!!!!
\(\left(\frac{1}{100.99}\right)-\left(\frac{1}{99.98}\right)-\left(\frac{1}{98.97}\right)-...-\left(\frac{1}{3.2}\right)-\left(\frac{1}{2.1}\right)\)
\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)
\(=\frac{1}{99}-\frac{1}{100}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1+\frac{1}{2}\right)\)
\(=\frac{1}{99}-\frac{1}{100}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{1}{100}-1+\frac{1}{99}\)
\(=\frac{2}{99}-\frac{101}{100}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-......-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(-\frac{1}{100.99}+\frac{1}{99.98}+...........+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(-\frac{1}{100}-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+......+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(-\frac{1}{100}-1\right)\)
\(=\frac{1}{100}+1\)
\(=\frac{101}{100}\)
a,=(1/3+3/5+1/15)+(3/4+-1/36)+(1/72-2/9)=1+26/36-15/72=1+(52-15)/72=1+37/72=109/72
b,=1/100-(1/1x2+1/2x3+...+1/97x98+1/98x99+1/99x100)
=1/100-(1/1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
=1/100-(1/1-1/100)=1/100-99/100=-98/100=-49/50
chỉ có mk mk giải thôi đó l-i-k-e đi
Đặt \(A=\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(-A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(-A=1-\frac{1}{99}\)
\(-A=\frac{98}{99}\)
\(A=\frac{-98}{99}\)
Chúc bạn học tốt ~
Đặt A = \(\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=> - A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
- A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
- A = \(1-\frac{1}{99}\)
- A = \(\frac{98}{99}\)
=> A = \(-\frac{98}{99}\)
Vậy A = \(-\frac{98}{99}\)
Hok tốt
Đặt A = \(\frac{1}{99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)
\(A=\frac{1}{99}-\left[-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{98.99}\right)\right]\)
\(A=\frac{1}{99}+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{99}\right)\)
\(A=\frac{1}{99}+\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{99}+\frac{98}{99}=1\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
Đặt E = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100}=\frac{99}{100}\)
\(50C=\left(\frac{1}{100}-\frac{99}{100}\right).5-=-\frac{49}{50}.50=-49\)
1. Để \(A_{min}\)thì \(x^4_{min}\)và \(2.x^2_{min}\) => \(x_{min}\) => \(x=0\)
Thay x vào ta có:\(A_{min}=0^4+2.0^2-7\)
\(A_{min}=0+0-7\)
\(A_{min}=-7\)
2. Ta có điểm M(1;5) => y=5;x=1
Thay x=1;y=5 vào ta có: \(5=a.1\)
=> a=5
4. Ta có: \(\frac{4x-9}{3x+y}-\frac{4y+9}{3y+x}=\frac{4x-\left(x-y\right)}{3x+y}-\frac{4y+\left(x-y\right)}{3y+x}\)
\(=\frac{4x-x+y}{3x+y}-\frac{4y+x-y}{3y+x}\)
\(=\frac{3x+y}{3x+y}-\frac{3y+x}{3y+x}\)
\(=1-1\)
\(=0\)
ban co bi gi ko lam thi phai cho mot it $ chu neu ko con lau ma lam cho
-49
mình giải rồi vòng 11 đó