-(127+38-21)-(121-27+48)+50

=-127-38+21-121+27-48+50

=(-127+27)-(38+48)+(21-121)+50

=-100-86-100+50

=-(100+100+86)+50

=-286+50

=-236

-39+(193-127+96)-(193+196-127)

=-39+193-127+96-193-196+127

=-39+(193-193)-(127+127)+(96-196)

=-39-100

=-139

\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)

\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)

\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)

\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)

=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)

Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)

=> x + 100 = 0

=> x = -100

Vậy x = -100

Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a

???                          

A=0

B=8

\(A=\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)

\(=\frac{25^{28}+25^{24}+...+25^0}{\left(25^{28}+25^{24}+...+25^0\right)+\left(25^{30}+23^{26}+...+25^2\right)}\)

\(=\frac{25^{28}+25^{24}+...+25^0}{\left(25^{28}+25^{24}+...+25^0\right)+25^2\left(25^{28}+23^{24}+...+25^0\right)}\)

\(=\frac{25^{28}+25^{24}+...+25^0}{\left(25^{28}+25^{24}+...+25^0\right)\left(1+25^2\right)}\)

\(=\frac{1}{1+25^2}\)

\(=\frac{1}{626}\)