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\(0< x< \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)
Mà \(x\in N\Rightarrow x=1\)
Ta có:
0<x<1/2+1/3+1/4
=> 0<x<13/12
=>0<x<1.083333....(1)
Mà x là số tự nhiên (2)
Từ (1) và (2) ta có x thuộc tập hợp rỗng

\(x+\frac{2}{5}=1-\frac{1}{6}\)
\(x+\frac{2}{5}=\frac{5}{6}\)
\(x=\frac{5}{6}-\frac{2}{5}=\frac{13}{30}\)
\(x=\frac{13}{30}\)

bài 1
Ta có : 2016/2017<1
2017/2018<1
Nên 2016/2017=2017/2018
Bài 1 :
a) Ta có : \(\frac{2016}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
Vì \(-\frac{1}{2017}< -\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\)
b) Ta có : \(\frac{2018}{2017}=1+\frac{1}{2017}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{2017}< \frac{1}{2016}\) nên \(\frac{2018}{2017}< \frac{2017}{2016}\)
Câu 2 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2015}{2016}\)
\(=\frac{1.2.3.4...2015}{2.3.4.5...2016}\)
\(=\frac{1}{2016}\)

\(\frac{1}{2016}\)+ \(\frac{3}{2016}\)+ \(\frac{5}{2016}\)+..........+ \(\frac{2015}{2016}\)= \(\frac{1+3+5+....+2015}{2016}\)
=\(\frac{1016064}{2016}\)= \(504\)
\(\frac{1}{2016}\)\(+\frac{3}{2016}\)\(+\frac{5}{2016}\)\(+...+\frac{2015}{2016}\)
\(=\frac{1+3+5+...+2015}{2016}\)
\(=\frac{1016064}{2016}\)
\(=504\)

1) \(\frac{4}{5}.\frac{8}{3}.\frac{x}{7}=\frac{96}{105}\)
\(\frac{32.x}{105}=\frac{96}{105}\)
\(32x=96\)
\(x=3\)
Bài 1: Tìm \(x\)
\(\frac45\times\frac83\times\) \(\frac{x}{7}\) = \(\frac{96}{105}\)
\(\frac{32}{15}\) x \(\frac{x}{7}\) = \(\frac{32}{35}\)
\(\frac{x}{7}\) = \(\frac{32}{35}\) : \(\frac{32}{15}\)
\(\frac{x}{7}\) = \(\frac{32}{35}\) x \(\frac{15}{32}\)
\(\frac{x}{7}\) = \(\frac37\)
\(x=\frac37\times7\)
\(x=3\)
b; \(\frac74\times\frac{3}{x}\times\) \(\frac{11}{5}\) = \(\frac{231}{200}\)
\(\frac74\) x \(\frac{3}{x}\) = \(\frac{231}{200}\) : \(\frac{11}{5}\)
\(\frac74\times\frac{3}{x}\) = \(\frac{231}{200}\) x \(\frac{5}{11}\)
\(\frac74\) x \(\frac{3}{x}\) = \(\frac{21}{40}\)
\(\frac{3}{x}\) = \(\frac{21}{40}\) : \(\frac74\)
\(\frac{3}{x}\) = \(\frac{21}{40}\) x \(\frac47\)
\(\frac{3}{x}\) = \(\frac{3}{10}\)
\(x=3:\frac{3}{10}\)
\(x=3\times\frac{3}{10}\)
\(x=10\)

1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
x = 0 ???
Đây là toán lớp 4?!