tìm x biet
\(\frac{1}{x}< 2va\frac{1}{x}>-3\)
Những câu hỏi liên quan
Tìm GTNN
A=x+y+\(\frac{1}{x}+\frac{1}{y}\)biet x+y=3
Xem thêm câu trả lời
tìm so nguyen x biet: a) \(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+..........+\frac{1}{\left(2x-1\right)x\left(2x+1\right)}=\frac{49}{99}\)
b) 1-3+32-33+.........+(-3)x=\(\frac{9^{1006}-1}{4}\)
tìm x, y thuộc Z , biet :
\(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
tim x biet
a.\(\frac{x+1}{10}+\frac{x+1}{11}=\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b.\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\)
a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
Đúng 0
Bình luận (0)
Tim x biet
\(-2\frac{1}{3}x-1\frac{3}{4}x+3\frac{2}{3}=3\frac{3}{5}\)
Tìm x biet:
a) 11(x - 6) = 4 x x + 11
b) \(4\frac{1}{3}\)(\(\frac{1}{6}\) - \(\frac{1}{2}\)) _< x _< \(\frac{2}{3}\) (\(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{3}{4}\)) với x C Z
Tim x biet:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\frac{\left(x+1-2\right)}{2.\left(x+1\right)}=\frac{2011}{4026}\)
Đúng 0
Bình luận (0)
Tim x biet
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)
=> \(1-\frac{2}{x+1}=\frac{2019}{2020}\)
=> \(\frac{2}{x+1}=\frac{1}{2020}=\frac{2}{4040}\)
=> x + 1 = 4040 => x = 4039
Tim x biet
c) \(x-3\frac{1}{2}x=-2\frac{6}{7}\)
d) \(-2\frac{1}{3}x-1\frac{3}{4}x+3\frac{2}{3}=3\frac{3}{5}\)