\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{3}{5}.2,75-1,2:\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}:\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}.\dfrac{11}{4}\)

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

\(=\dfrac{11}{4}.0\)

\(=0\)

\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{2}{5}.2,75-\left(-1,2\right):\dfrac{4}{11}\)

\(=\dfrac{11}{4}.\dfrac{2}{5}+\dfrac{7}{5}.\dfrac{11}{4}+\dfrac{6}{5}.\dfrac{11}{4}\)

\(=\dfrac{11}{4}.\left(\dfrac{2}{5}+\dfrac{7}{5}+\dfrac{6}{5}\right)\)

\(=\dfrac{11}{4}.\dfrac{15}{5}\)

\(=\dfrac{11}{4}.3\)

\(=\dfrac{33}{4}\)

\(=1.1-4.4-3.3\)

\(=-6.6\)

-11/5 có cần cách giải ko? Nhân tiện chúc bn năm ms zui zẻ nhé!!!!!!!!!!!!!!!!!!!!

Mik tính đc -6,6

Bài 1: 

a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)

\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)

\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)

\(=-10\cdot1.7=-17\)

b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)

\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)

\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)

\(=-\dfrac{539}{60}\)

c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)

\(=10\)

dấu -. là sao  ?

bo dau - di nhe

Mà cái này chẳng phải tính thuận tiện đâu, cứ từ từ mà tính