11 tháng 2 2018 lúc 14:05
Các câu hỏi tương tự
tìm x
\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+\left|x+\frac{1}{3.4}\right|+...+\left|x+\frac{1}{99.100}\right|\)=100x
TÌM x BIẾT\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
Tìm x , biết
a) \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)
b) \(|x+\frac{1}{1.2}|+|x+\frac{1}{2.3}|+|x+\frac{1}{3.4}|+.....+|x+\frac{1}{99.100}|=100x\)
\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+\left|x+\frac{1}{3.4}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
Tìm x bt:
\(\left[x+\frac{1}{1.2}\right] +\left[x+\frac{1}{2.3}\right]+\left[x+\frac{1}{3.4}\right]+...+\left[x+\frac{1}{99.100}\right]=100x\)
cái [] là trị tuyệt đối nhé
Tìm x biết
|\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+\left|x+\frac{1}{3.4}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
Gỉai nhanh giúp mình nha mn. Cảm ơn trước nha
a)Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{99.100} 1b)Bfrac{1}{3}+left(frac{1}{3}right)^2+left(frac{1}{3}right)^3+...+left(frac{1}{3}right)^{100} frac{1}{2}c)frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{49.50}frac{1}{26}+frac{1}{27}+...+frac{1}{50}d)Afrac{1}{1.2}+frac{1}{3.4}+...+frac{1}{99.100}.CMRfrac{7}{12} A frac{5}{6}AI ĐÚNG MINK left(TICKright)CHO (làm đc trên 2 câu)
Đọc tiếp
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
b)B=\(\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{100}< \frac{1}{2}\)
c)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
d)A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}.CMR\frac{7}{12}< A< \frac{5}{6}\)
AI ĐÚNG MINK \(\left(TICK\right)\)CHO (làm đc trên 2 câu)
\(B=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)....\left(1-\frac{2}{99.100}\right)\)
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)......\left(1-\frac{2}{99.100}\right)\)