Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mình chỉ làm cho bạn câu d và e thôi
d) ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +....... +1/99 - 1/100 ) . (x - 3)=1
( 1 - 1/100 ) . (x - 3 )=1
99/100.(x -3)=1
x - 3 = 1:99/100
x - 3 =100/99
x = 100/99 + 3
x = 397/99
e) (1/2 . (1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +.....+1/99 - 1/101 ) . (x+2) =3/101
(1/2 . ( 1 - 1/101 ).(x+2)=3/101
(1/2 . 100/101 ) . (x + 2) =3/101
100/202 . ( x + 2 )= 3/101
50/101 . (x + 2 ) = 3/101
x + 2 = 3/101 :50/101
x+2=3/50
x =3/50-2
x= -97/100
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Ủng hộ mk nha !!! ^_^
Ta có: \(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)
\(=2-\frac{1}{100!}< 2\)
\(B=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(B=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(B=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
2Q = 1-1/3-1/2+1/4+1/3-1/5-1/4+1/6-........+1/97-1/99-1/98+1/100 = 1-1/2-1/99+1/100 = 4949/9900 >> Q = 49499/19800
\(Q=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}+\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{99}{100}=\frac{99}{200}\) (không chắc cho lắm :v)
\(D=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
Làm tắt nha :
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{98}{100}\)
\(D=\frac{1}{2}\left(\frac{98}{99}-\frac{98}{100}\right)\)
Tự tính nốt nha