=202*6,25+202*2,5+202*1,25+1

=2020+1

=2021

{ 2019 - [ -521]} - {202 + 521} + {-456 + 202}

=  2019 + 521 - 202 - 521 + -456 + 202

= 2019 + (521 - 521) + (-202 + 202) + -456

= 2019 + 0 + 0 + -456

= 2019 + -456

= 1563

[2019 - (-521)] - (202 + 521) + (-456 + 202)

= 2019 + 521 - 202 - 521 - 456 + 202

= (2019 - 456) + (521 - 521) - (202 - 202)

= 1563

= (2019+521) - (202+521) +(456+202)

= 2019 + 521 -202 - 521 +456 +202 =2019 +456 = 2475

      \(202x\frac{3}{4}+202x\frac{5}{4}+202x8\)

\(=202x\left(\frac{3}{4}+\frac{5}{4}+8\right)\)

\(=202x\left(\frac{8}{4}+8\right)\)

\(=202x\left(2+8\right)\)

\(=202x10\)

\(=2020\)

\(=202*(\frac{3}{4}+ \frac{5}{4}+8) = 202*9 = 1818\)

    202 x \(\frac{3}{4}\) + 202 x\(\frac{5}{4}\) + 202 x 8

= 202 x ( \(\frac{3}{4}\)+\(\frac{5}{4}\)+ 8 )

= 202 x ( \(\frac{8}{4}\)+ 8 )

= 202 x ( 2 + 8 )

= 202 x 10

= 2020 . 

=1+(2-3-4+5)+3-4-5+6)+...+(200-201-202+203)+204

=1+0+0+...+0+204

=1+204

=205

tinh 

1+2 -3 -4 +5+3-4-5+6 +4-5-6 +7 + ....+200-201 -202 + 203 + 204 =1+0+...+0+204=1+204=205.

bn k cho mik nha. ^-^ thanks bn trc.

ban oi minh chua hoc so am 

S= 2-1-2-3-4-....-201-202 

  = 2-(1+2+3+4+....+202)

=2- (202:2)( 202+1)

= 2- 101.203

Ta có :

\(202^{303}=\left(202^3\right)^{101}=8242408^{101}\)

\(303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(\Rightarrow202^{303}>303^{202}\)

202^303=(202^3)^101=8242408^101

303^202=(303^2)^101=91809^101

vì 8242408>91809 nên 8242408^101>91809^101

=>202^303>303^202

ta có:

202³⁰³=(202³)¹⁰¹=8242408¹⁰¹

303²⁰²=(303²)¹⁰¹=91809¹⁰¹

vì 8242408>91809 =>8242408¹⁰¹>91809¹⁰¹ hay 202³⁰³>303²⁰²

Please help me!!!

=-(202+1)-(201+2)-......-(1+202)

=-203-203-203-......-203

=-20503

-20503 đó bn

\(\frac{201}{202}+\frac{202}{205}\)Và \(201+\frac{202}{202}+205\)

\(=\frac{201}{202}=\frac{201}{202}+\frac{1}{202}=\frac{202}{202}\)

\(\frac{202}{205}=\frac{202}{205}+\frac{3}{205}=\frac{205}{205}\)

\(201+1+205\)

Vậy \(1+1=2\)và \(407\)

=> \(\frac{201}{202}+\frac{202}{205}< 201+\frac{202}{202}+205\)

Ta có: \(\frac{201+202}{202+205}=\frac{201}{202+205}+\frac{202}{202+205}\)

Ta có: 202<202+205 => \(\frac{201}{202}>\frac{201}{202+205}\)(1)

205<202+205 => \(\frac{202}{205}>\frac{202}{202+205}\)(2)

Từ (1) và (2) => \(\frac{201}{202}+\frac{202}{205}>\frac{201+202}{202+205}\)