Câu đầu em xem lại đề bài sao có hai dấu bằng.

Câu 2: 

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

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

y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)

y \(\times\) \(\dfrac{7}{4}\)            = \(\dfrac{4}{5}\)

y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)

y = \(\dfrac{16}{35}\)

\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{16\times10}{5\times11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{32+13}{11}\\ =\dfrac{45}{11}\)

\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16\times3}{3\times5}-\dfrac{6}{5}\\ =\dfrac{48}{15}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{16-6}{5}\\ =\dfrac{10}{5}\\ =2\)

\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{3\times5+1}{5}\times\dfrac{10}{11}+\dfrac{1\times11+2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{45}{11}\)

\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{5\times3+1}{3}:\dfrac{1\times3+2}{3}-\dfrac{1\times5+1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{10}{5}=2\)

\(\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)

\(=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8})}{5\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{8}\right)}\)

\(=\dfrac{3}{5}+\dfrac{2}{5}\)

\(=1\)

Bài 1: 

a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)

c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)

31​−53​+75​−97​+119​−1311​+1513​+1311​−119​+97​−75​+53​−31​
=(13−13)−(35−35)+(57−57)−(79−79)+(911−911)−(1113−1113)+1315=(31​−31​)−(53​−53​)+(75​−75​)−(97​−97​)+(119​−119​)−(1311​−1311​)+1513​
=1315=1513​

 

Đề bài yêu cầu gì?

/:)))

???

Jjj

a) \(\dfrac{3}{5}:\left(\dfrac{7}{11}+\dfrac{4}{11}\right)=\dfrac{3}{5}.1=\dfrac{3}{5}\)

 c2: \(\dfrac{7}{11}:\dfrac{3}{5}+\dfrac{4}{11}:\dfrac{3}{5}=\dfrac{7}{11}.\dfrac{5}{3}+\dfrac{4}{11}.\dfrac{5}{3}=\dfrac{3}{5}\)

a) 

C1: \(\dfrac{7}{11}:\dfrac{3}{5}+\dfrac{4}{11}:\dfrac{3}{5}=\left(\dfrac{7}{11}+\dfrac{4}{11}\right):\dfrac{3}{5}=1:\dfrac{3}{5}=\dfrac{5}{3}\)

\(\dfrac{6}{11}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{5}{11}\)

\(=\dfrac{3}{7}.\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\)

\(=\dfrac{3}{7}.1=\dfrac{3}{7}\)

1.4x26x5.6=203.84

\(\dfrac{6}{11}x\dfrac{3}{7}+\dfrac{3}{7}x\dfrac{5}{11}=\dfrac{3}{7}x\left(\dfrac{6}{11}+\dfrac{5}{11}\right)=\dfrac{3}{7}x1=\dfrac{3}{7}\)

\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)

1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.

2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.

3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).