1/3-(-2/7+1/3)
= 1/3 + 2/7 - 1/3
= (1/3 - 1/3) + 2/7
= 2/7

\(\frac{1}{2}\cdot\frac{18}{14}+\frac{1}{2}\cdot\frac{3}{14}-\frac{1}{2}\cdot\frac{7}{14}\)

\(=\frac{1}{2}\left(\frac{18}{14}+\frac{3}{14}-\frac{7}{14}\right)\)

\(=\frac{1}{2}\cdot1=\frac{1}{2}\)

\(a:=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}\right)\\ =1+\dfrac{3}{5}=1\dfrac{3}{5}\\ b:=\left(\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{5}{14}+\dfrac{3}{14}\right)\\ =\dfrac{3}{7}+\dfrac{8}{14}\\ =\dfrac{3}{7}+\dfrac{4}{7}=1\)

a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)

Vậy A<\(\frac{1}{2}\).

b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                  \(\frac{1}{3^2}< \frac{1}{2.3}\)

                  \(\frac{1}{4^2}< \frac{1}{3.4}\)

                    ...

                   \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

Vậy \(B< 1\).

Câu 1: 

\(\dfrac{2}{5}-\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{8}{20}-\dfrac{5}{20}+\dfrac{6}{20}=\dfrac{8-5+6}{20}=\dfrac{9}{20}\) 

Câu 2:

\(\dfrac{-2}{5}:\left(1-\dfrac{1}{10}\right)=\dfrac{-2}{5}:\dfrac{9}{10}=\dfrac{-2}{5}.\dfrac{10}{9}=\dfrac{-2.10}{5.9}=\dfrac{-20}{45}=\dfrac{-4}{9}\) 

Câu 3:

\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}=\dfrac{7}{18}+\dfrac{1}{5}=\dfrac{53}{90}\) 

Câu 4:

\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\) 

\(=\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\) 

\(=\dfrac{2}{7}.1\) 

\(=\dfrac{2}{7}\)

Câu 1

\(\dfrac{2}{5}\)-\(\dfrac{1}{4}\)+\(\dfrac{3}{10}\)= \(\dfrac{8}{20}\)-\(\dfrac{5}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{3}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{9}{20}\)

Câu 2

-\(\dfrac{2}{5}\):(1-\(\dfrac{1}{10}\))= -\(\dfrac{2}{5}\):\(\dfrac{9}{10}\)=-\(\dfrac{2}{5}\).\(\dfrac{10}{9}\)=-\(\dfrac{4}{9}\)

Câu 3

\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}\)= \(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}.\dfrac{14}{5}\)=\(\dfrac{7.4}{4.2.9}+\dfrac{1.14}{14.5}\)=\(\dfrac{7}{18}+\dfrac{1}{5}\)=\(\dfrac{35}{90}+\dfrac{18}{90}\)=\(\dfrac{53}{90}\)

Câu 4

\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)=\(\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)=\(\dfrac{2}{7}.1\)=\(\dfrac{2}{7}\)

a: \(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{3}\cdot\dfrac{1}{2}=\dfrac{2}{5}+\dfrac{10}{5}\cdot\dfrac{1}{2}=\dfrac{2}{5}+1=\dfrac{7}{5}\)

A=\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)+\(\frac{1}{132}\)

A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)

A= \(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)

A= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)

Cho phản ứng xFeS2 + yH2SO4 (đặc, nóng) -> zFe(SO4)3 + tSO2 + aH2O. Các giá trị a ; x ; y ; t ; z lần lượt là:
A. 2 ; 14 ; 15 ; 1 ; 14

B. 1 ; 14 ; 14 ; 15 ; 2

C. 15 ; 2 ; 14 ; 14 ; 1

D. 14 ; 2 ; 14 ; 15 ; 1