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Bài 1:
a) (2x-3). (x+1) < 0
=>2x-3 và x+1 ngược dấu
Mà 2x-3<x+1 với mọi x
\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)
b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu
Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)
Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
=>....
Bài 2:
\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\cdot\frac{998}{3003}\)
\(=\frac{499}{3003}\)
\(S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
Đặt tên bthuc là A
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(2A=1-\frac{1}{21}=\frac{20}{21}\)
=>\(A=\frac{20}{21}:2=\frac{10}{21}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)
\(=\frac{9}{19}\)
A = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
2A = 2 . \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
2A = \(\frac{1}{3}-\frac{1}{99}\)
2A = \(\frac{32}{99}\)
A = \(\frac{32}{99}\div2\)
A =\(\frac{16}{99}\)
_HT_
tớ làm câu b thôi, câu a nhân 1/2 lên là đc
\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2.n-1}-\frac{1}{2n+1}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)
p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)
X+(1/1.3+1/3.5+1/5.7+...+1/99.101)=100
X+(2/1.3+2/3.5+2/5.7+...+2/99.101)=100
X+(1 -1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)=100
X+(1-1/101)=100
X+100/101=100
X=100-100/101
X=10000/101
gọi biểu thức là A
ta có :
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{19.21}\)
=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}...\frac{2}{19.21}\)
2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{21}\)
2A = 1 - \(\frac{1}{21}\)
2A = \(\frac{20}{21}\)
A = \(\frac{20}{21}:2=\frac{10}{21}\)
=1-(1/3.5+1/3.7+1//7.9+...+1/55.57)
=1-1/2.(2/3.5+2/5.7+2/7.9+...+2/55.57)
=1-1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/55-1/57)
=1-1/2(1/3-1/57)
=1-1/2.18/57
=1-9/57
=48/57
=
1-(1/3.5+1/5.7+1/7.9+....+1/53.55+1/55.57)
=1-1/2.[1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57]
=1-1/2.[1/3-1/57]
=1-1/2.54/171
=1-28/171
=143/171.
\(S2=\left(\frac{1}{3}-\frac{1}{1001}\right):2=\frac{499}{3003}\)
\(S2=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{999.1001}\)
\(S2=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(S2=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(S2=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(S2=\frac{1}{2}.\frac{998}{3003}=\frac{499}{3003}\)