bài 1
a,-1/8 - (320% + 9/5 ) . (1/4 - 1/8
b,4/3.5 + 4/5.7 + 4/7.9 + ......+4/99.100
bài 2
a,x+ 3/22 = 27/121.11/9
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a,[(4x+28):3+55]:5=35
(4x+28):3+55=175
(4x+28):3=120
4x+28=360
4x=332
x=83
a) [( 4x + 28 ) : 3 + 55] : 5 = 35
( 4x + 28 ) : 3 + 55 = 175
( 4x + 28 ) : 3 = 120
4x + 28 = 360
4x = 332
x = 83
b) \(\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\right)\cdot x=\frac{9}{7}\)
\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\cdot x=\frac{9}{7}\)
\(\left(\frac{1}{3}-\frac{1}{21}\right)\cdot x=\frac{9}{7}\)
\(\frac{2}{7}\cdot x=\frac{9}{7}\)
\(x=\frac{9}{2}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)
G=\(\frac{3}{2.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{2015.2017}\)
G=\(3.\left(\frac{1}{2.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\right)\)
G=\(3.\left(\frac{1}{2}.\frac{1}{5}+\frac{1}{5}.\frac{1}{7}+\frac{1}{7}.\frac{1}{9}+...+\frac{1}{2013}.\frac{1}{2015}+\frac{1}{2015}.\frac{1}{2017}\right)\)
G=\(3.\left(\frac{1}{2}+\frac{1}{2017}\right)\)
G=1.5
Anh ko bik có đúng ko nữa lâu quá rồi. Em thông cảm nhé
thôi chịu nhiều quá ai mà làm đc tự đi mà làm hỏi thì hỏi thì hỏi ít thôi người ta còn trả lời đc .
Đặt \(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{3}-\frac{1}{101}\)
\(\Rightarrow\frac{1}{2}A=\frac{101}{303}-\frac{3}{303}\)
\(\Rightarrow\frac{1}{2}A=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}\div\frac{1}{2}\)
\(\Rightarrow A=\frac{199}{303}\)
\(x+\frac{3}{22}=\frac{27}{121}.\frac{11}{9}\)
\(\Leftrightarrow x+\frac{3}{22}=\frac{27.11}{121.9}\)
\(\Leftrightarrow x+\frac{3}{22}=\frac{3.1}{11.1}\)
\(\Leftrightarrow x+\frac{3}{22}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}-\frac{3}{22}\)
\(\Leftrightarrow x=\frac{6}{22}-\frac{3}{22}\)
\(\Leftrightarrow x=\frac{3}{22}\)