1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)

\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)

2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)

\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)

3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)

\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)

1.   \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)

\(\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{4}{3}+\dfrac{19}{3}=\dfrac{1+2+4+19}{3}=\dfrac{26}{3}\)

\(\dfrac{3}{4}+\dfrac{4}{4}+\dfrac{5}{4}+\dfrac{6}{4}+\dfrac{x}{4}+\dfrac{8}{2}+\dfrac{9}{4}\)

=\(\dfrac{3}{4}+\dfrac{4}{4}+\dfrac{5}{4}+\dfrac{6}{4}+\dfrac{x}{4}+\dfrac{16}{4}+\dfrac{9}{4}\)

=\(\dfrac{3+4+5+6+x+16+9}{4}=\dfrac{43+x}{4}\)

Cảm ơn và chúc Lê Minh Quang học tốt nhé!

Mình đã tick rùi nha

Thanks

1/3+2/3+4/3+19/3=26/3
3/4+4/4+5/4+6/4+x/2+8/2+9/4=43+x/4

a

= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}

Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.

Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .

Kết quả là : C1=\(\dfrac{2}{3}\)

[] cai dau nay la gia tri tuyet doi nha

\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)

\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)

\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)

\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)

a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)

\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)

\(=\dfrac{2}{5}+-2\)

\(=\dfrac{2}{5}+\dfrac{-10}{5}\)

\(=\dfrac{-8}{5}\)

a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{5}+-2=\dfrac{-8}{5}\)  

b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)=\dfrac{19}{5}-\dfrac{9}{4}-\dfrac{9}{5}=\left(\dfrac{19}{5}-\dfrac{9}{5}\right)-\dfrac{9}{4}=2-\dfrac{9}{4}=\dfrac{-1}{4}\) 

c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\) 

\(=\dfrac{-3}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\) 

\(=\dfrac{-3}{5}.1+\dfrac{3}{5}\) 

\(=\dfrac{-3}{5}+\dfrac{3}{5}\) 

\(=0\) 

d) \(40\%-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\) 

\(=\dfrac{2}{5}-\dfrac{12}{7}:\dfrac{3}{7}+\dfrac{9}{5}\) 

\(=\dfrac{2}{5}-4+\dfrac{9}{5}\) 

\(=\dfrac{-9}{5}\)

\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...

Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)

\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)

\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)