Đề bài lạ thế!

\(A=-\frac{8}{5}x^3+\frac{36}{5}x^2y-\frac{54}{5}xy^2+\frac{27}{5}y^3\)

\(=-\frac{1}{5}\left(8x^3-36x^2y+54xy^2-27y^3\right)\)

=\(-\frac{1}{5}\left(\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\right)\)

\(=-\frac{1}{5}\left(2x-3y\right)^3=-\frac{1}{5}.4^3=-\frac{64}{5}\)

-2 2/3= -8/3

 a bằng:

36/5 : (-8/3)= -27/10

A bằng: -27/10

a) Với \(\frac{m}{n} = \frac{{ - 5}}{6}\), giá trị của biểu thức là:

\(\begin{array}{l}A = \frac{{ - 2}}{3} - \left( {\frac{{ - 5}}{6} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - \frac{{-20}}{6}.\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - \frac{{ 25}}{{12}}\\A = \frac{{ - 33}}{{12}}\end{array}\)

b) Với \(\frac{m}{n} = \frac{5}{2}\) , giá trị của biểu thức là:

\(\begin{array}{l}A = \frac{{ - 2}}{3} - \left( {\frac{5}{2} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - 0.\frac{{ - 5}}{8} = \frac{{ - 2}}{3}\end{array}\)

c) Với \(\frac{m}{n} = \frac{2}{{ - 5}}\) , giá trị của biểu thức là:

\(\begin{array}{l}A = \frac{-2}{3} - \left( {\frac{2}{{ - 5}} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \left( {\frac{{ - 4}}{{10}} + \frac{{ - 25}}{{10}}} \right).\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \frac{{ - 29}}{{10}}.\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \frac{{29}}{{16}}\\A = \frac{{-32}}{{48}} - \frac{{87}}{{48}}\\A = \frac{{ - 119}}{{48}}\end{array}\).

c1 la :-2.7

c2 la :-3.1

c3 la :1.3333

Câu 1: -2,7

Câu 2: 2016

Câu 3: \(\frac{7}{15}\)

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)