(91-99+98)-(-99+98)=91-99+98+99-98=91

(99-98+97)-(99+97+98)=99-98+97-99-97-98=(-98)*2=-196

1. (91 - 99 + 98) - (-99 + 98)

= 91 - 99 + 98 + 99 - 98

= (-99 + 99) + (98 - 98) + 91

= 0 + 0 + 91

= 91

2. (99 - 98 + 97) - (99 + 97 + 98)

= 99 - 98 + 97 - 99 - 97 - 98

= (99 - 99) + (97 - 97) + (-98 - 98)

= 0 + 0 + (-196)

= -196

\(1+\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}\)

\(=1+1+\frac{1}{98}-\left(1+\frac{1}{97}\right)+\frac{1}{97}-\frac{1}{98}\)

\(=1+1+\frac{1}{98}-1-\frac{1}{97}+\frac{1}{97}-\frac{1}{98}\)

\(=1+1-1\)

\(=1\)

thank

tính B

đề :

= 1/100 - (1 / 100.99 +1/99.98 + ...+ 1/3.2 +1/2.1 ) 

=1/100 - (1 /1.2 +1/ 2.3 +...+ 1/ 98.99 +1 / 99.100)

=1/100 -( 1- 1/ 2 +1/2 -1/3 +...+1/98 -1/99 +1/99 -1/100)

=1/100 - ( 1- 1/100)

=1/100 - 99 /100

= -98/100

= -49 /50

Ta có: \(A=\frac{97^{98}+1}{97^{99}+1}\Rightarrow97A=\frac{97^{99}+97}{97^{99}+1}=\frac{97^{99}+1+96}{97^{99}+1}=1+\frac{96}{97^{99}+1}\)

\(B=\frac{97^{97}+1}{97^{98}+1}\Rightarrow97B=\frac{97^{98}+97}{97^{98}+1}=\frac{97^{98}+1+96}{97^{98}+1}=1+\frac{96}{97^{98}+1}\)

Vì \(\frac{96}{97^{99}+1}< \frac{96}{97^{98}+1}\Rightarrow1+\frac{96}{97^{99}+1}< 1+\frac{96}{97^{98}+1}\Rightarrow97A< 97B\Rightarrow A< B\)

Vậy A < B

Ta thấy A < 1 và 96 > 1 nên ta có:

            A <     97⁹⁸ + 1 + 96 / 97⁹⁹ + 1 + 96

      =>  A <     97⁹⁸ + 97 / 97⁹⁹ + 97

      =>  A <     97(97⁹⁷ + 1) / 97(97⁹⁸ + 1)

      =>  A <     97⁹⁷ + 1 / 97⁹⁸ + 1 = B

      => A < B

A<B

nhanh và gọn, lẹ

(-99)+ (-98) + (-97) +... + 97 + 98 + 99 + 100

= (-99 +99) + (-98 + 98) + (-97+97) +... + (-2+2) + (-1 + 1) + 100

= 0 + 0 + 0 + ... + 0 + 0 + 100

=100

=[(-99)+99]+[(-98)+98]+[(-97)+97]+..+100

=0+0+0+...+100=100