Theo đề ta có:

\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)\)\(-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)

= \(5+\)\(\frac{1}{5}-\frac{2}{9}\)-\(2+\frac{1}{23}+2+\frac{3}{5}+\frac{5}{6}-8+\frac{2}{3}-\frac{1}{18}\)

=\(\left(5+2-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{5}{6}-\frac{2}{3}+\frac{1}{18}\right)+\frac{1}{23}\)

=  -1 +\(\frac{4}{5}\)\(-\frac{-11}{9}\)+\(\frac{1}{23}\)

= -1 +\(\frac{4}{5}+\frac{11}{9}+\frac{1}{23}\)

\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{5}+\frac{5}{6}\right)-\left(8-\frac{2}{3}-\frac{1}{18}\right)\)

= \(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+2+\frac{3}{5}-\frac{5}{6}-8+\frac{2}{3}+\frac{1}{18}\)

= \(\left(5-8\right)+\left(\frac{1}{5}+\frac{3}{5}\right)-\left(\frac{2}{9}-\frac{1}{18}-\frac{2}{3}\right)-\left(2-2\right)+\frac{1}{23}-\frac{5}{6}\)

= \(\left(-3\right)+\frac{4}{5}+\frac{1}{2}+\frac{1}{23}-\frac{5}{6}\)

= \(\left(\left(-3\right)+\frac{4}{5}+\frac{1}{2}-\frac{5}{6}\right)+\frac{1}{23}\)

= \(-\frac{38}{15}+\frac{1}{23}\)

= \(-\frac{859}{345}\)

-29 

256

làm cụ thể ra hộ mình

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

b)=(1.2.3....8).(9-1-8)

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

Câu 1:

\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4

Vậy A = 4

Câu 2:

\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)

\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)

\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)

\(B=1000-3.\left\{400-273\right\}\)

\(B=1000-3.127\)

\(B=1000-381\)

\(B=619\)

Vậy B = 619

A = -2 nhé , 

gợi ý các bạn chọn (k) đúng cho mình.