tìm x biết: \(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+.....+\left|x+\dfrac{1}{97.99}\right|=50x\)
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28 tháng 3 2017
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
22 tháng 3 2021
\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)
\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)
\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)
\(\Rightarrow32x+32=31x+62\)
\(\Rightarrow x=30\)
Vậy x=30
Chúc bn học tốt
23 tháng 11 2021
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
15 tháng 3 2018
Mấy câu này dễ mà,động não lên chứ bạn:v
Link______________Link
h) \(\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)
\(\ge\left|x-1+3-x\right|=2\)
\(\Rightarrow x+1>2\Leftrightarrow x>1\)
Vậy: \(\left\{{}\begin{matrix}x>1\\x\in R\end{matrix}\right.\)
Câu b xét khoảng tương tự với cái link t đưa thôi
14 tháng 3 2018
hơi bức xúc rồi đó
tau chỉ muốn kiểm tra lại thôi
21 tháng 8 2021
Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
Với mọi x ta có :
+) \(\left|x+\dfrac{1}{1.3}\right|\ge0; \)
+) \(\left|x+\dfrac{1}{3.5}\right|\ge0;\)
.....................................
+) \(\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+.......+\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Leftrightarrow50x\ge0\)
\(\Leftrightarrow x\ge0\)
Khi \(x\ge0\) ta được :
+) \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3}\)
+) \(\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5}\)
.............................................
+) \(\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\)
\(\Leftrightarrow\left(x+\dfrac{1}{1.3}\right)+\left(x+\dfrac{1}{3.5}\right)+......+\left(x+\dfrac{1}{97.99}\right)=50x\)
\(\Leftrightarrow49x+\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow x=\dfrac{16}{99}\)
Vậy...