ĐK:`x>=1/2;y>=1`

TH1:`\sqrt{2x-1}+\sqrt{y-1}=0<=>x=1/2;y=1=>x=2y`

TH2:`\sqrt{2x-1}+\sqrt{y-1}>0`

GT`<=>2x-y=\sqrt{y-1}-\sqrt{2x-1}`

`<=>2x-y=(y-1-(2x-1))/(\sqrt{2x-1}+\sqrt{y-1})`

`<=>(2x-y)(1+1/(\sqrt{2x-1}+\sqrt{y-1}))=0`

Dễ thấy biểu thức trong ngoặc luôn dương

`=>2x-y=0<=>2x=y`

Trong mọi TH ta luôn có `2x=y`

a,

`3A=3+3^3+3^3+...+3^{53}`

`3A-A=(3+3^3+3^3+...+3^{53})-(1+3+3^3+3^3+...+3^{52})`

`2A=3^{53}-1`

`A=(3^{53}-1)/2`

b,

`A=1+3+3^3+3^3+...+3^{52}`

`A=(1+3+3^2)+(3^3+3^4+3^5)+....+(3^{50}+3^{51}+3^{52})`

`A=(1+3+3^2)+3^3*(1+3+3^2)+....+3^{50}*(1+3+3^2)`

`A=(1+3+3^2)*(1+3^3+....+3^{50})`

`A=13*(1+3^3+....+3^{50})`

Do `13 \vdots 13 => A=13*(1+3^3+....+3^{50})\vdots 13 `

Vậy `A \vdots 13 `

`(x+2)(x+3)-2 \sqrt{x^2+5x+3}=6(x >=-1` hoăc `x<=-3/2)`

`<=> ( x^2+5x+6)-2 \sqrt{x^2+5x+3}=6`

Đặt `x^2+5x+3=t(t>=0)`

`<=> t+3-2 \sqrt{t}-6=0`

`<=> t-2\sqrt{t}-3=0`

`<=> ( \sqrt{t}-3)(\sqrt{t}+1)=0`

Do `t>=0=> \sqrt{t}+1>0`

`=> \sqrt{t}-3=0`

`<=> \sqrt{t}=3`

`<=>t=9`

Với `t=9<=>x^2+5x+3=9`

`<=> x^2+5x-6=0`

`<=>(x-1)(x+6)=0`

`<=> [(x-1=0),(x+6=0):}`

`<=> [(x=1(tm)),(x=-6(tm)):}` 

Vậy phương trình có tập nghiệm `S={1;-6}`

Ta có `:`

`a = 2013 xx 2013`

`= 2013 xx ( 2017-4)`

`=2013xx2017-4xx2013`

`=(2009+4)xx2017-4xx2013`

`=2017xx2009+4xx2017-4xx2013`

`= 2017xx2009 +4xx(2017-2013)`

`= 2017xx2009 + 4xx4`

`= 2017xx2009+16`

Do `2017xx2009+16` `> 2017 xx 2009` nên `a > b`