Question

cho B= 1\ 2(n-1)^2 +3 để tìm GTLN

a, x+2\327 + x+3\326 + x+4\325 + x+5\324 + x+349\5 =0

b, 2a+9\a+3  +  5a+17\a+3 - 3a\a+3 là số nguyên

Answer 1

a.

Ta có:

(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+349)/5=0

<=>(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+329)-4   (giải thích: (x+349)/5=(x+329+20)/5=(x+329)/5+4)

<=>1+(x+2)/327+1+(x+3)/326+1+(x+4)/325+1+(x+5)324+(x+329)/5=0

<=>(x+329)/327+(x+329)/326+(x+329)/325+(x+329)/324+(x+329)/5=0

<=>x+329(1/327+1/326+1/325+1/324+1/5)=0

Vì (1/327+...+1/5) khác 0 => x+329=0

=>x=-329