`|2x-1|-|3x-4|=0`

`|2x-1|=|3x-4|`

`[(2x-1=3x-4),(2x-1=-3x+4):}`

`[(x=3),(x=1):}`

`(3x+1)^2-9x(x+2)=x-1`

`<=>9x^2+6x+1-9x^2-18x=x-1`

`<=>-12x+1=x-1`

`<=>-13x=-2`

`<=>x=2/13`

`x^2-4x+3>=0`

`<=>(x-3)(x-1)>=0`

TH1: `{(x-3>=0),(x-1>=0):} <=> {(x>=3),(x>=1):} <=>x>=3`

TH2: `{(x-3<=0),(x-1<=0):}<=>{(x<=3),(x<=1):}<=>x<=1`

Vậy `x>=3 ; x<=1`.

`(3\sqrt2-2\sqrt3)/(\sqrt3-\sqrt2) - 5/(1+\sqrt6)`

`= (\sqrt6 (\sqrt3-\sqrt2))/(\sqrt3-\sqrt2) - (5(1-\sqrt6))/(1^2-\sqrt6^2)`

`=\sqrt6 - (5(1-\sqrt6))/(-5)`

`=\sqrt6 +1-\sqrt6`

`=1`

`=2\sqrt6-1`

`(2x)^2=2^2 . x^2 = 4.x^2`

a) `6/18+ (-14)/21 = 1/3 - 2/3=-1/3`

b) `3/5 - (-1/2) = 3/5+1/2=11/10`

c) `2/7 : 3/4 = 2/7 . 4/3 = 8/21`

d) `(-28)/33 . (-3)/4 = 28/33 . 3/4=7/11`

`(x+y+z+t)(x+y-z-t)`

`=[(x+y)+(z+t)][(x+y)-(z+t)]`

`=(x+y)^2-(z-t)^2`

`=(x+y)^2+[-(z-t)^2]`

1. Vì "which" thay thế cho "tokbokki" - "Korean food".

2.  Câu cần trạng ngữ -> loại A, C

Phía trước là "in July" = thời gian -> chọn D.

`\sqrt(x^2)=15`

`<=>|x|=15`

`<=> [(x=15),(x=-15):}`