Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

`|5x-2|=|7x-3|`

Trường hợp 1 :

`5x-2=7x-3`

`<=>2x=1`

`<=>x=1/2`

Trường hợp 2 :

`5x-2=-(7x-3)`

`<=>5x-2=-7x+3`

`<=>12x=5`

`<=>x=5/12`

Vậy `S={1/2;5/12}`

`|3x-1|=|2x+5|`

Trường hợp 1 :

`3x-1=2x+5`

`<=>x=6`

Trường hợp 2 :

`-(3x-1)=2x+5`

`<=>-3x+1=2x+5`

`<=>-5x=4`

`<=>x=-4/5`

Vậy `S={6;-4/5}`

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

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

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

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

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Có : `x^2-y^2=15`

`->(x-y)(x+y)=15`

`->5(x+y)=15`

`->x+y=3`

Mà `x-y=5`

`->x=4;y=-1`

`->x^3+y^3=4^3-(-1)^3=64+1=65`

`x^4-2x^3+10x^2-20x=0`

`<=>x^3(x-2)+10x(x-2)=0`

`<=>(x^3+10x)(x-2)=0`

`<=>x(x^2+10)(x-2)=0`

`<=>`$\left[\begin{matrix} x=0\\ x^2+10=0\\x-2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=0\\ x^2=-10 \ \rm(loại) \\x=2\end{matrix}\right.$

Vậy `S={0;2}`

`3x^2-30x+73`

`=3(x^2-10x+73/3)`

`=3(x^2-2.x.5+25-2/3)`

`=3(x^2-2.x.5+5^2)-3 . 2/3`

`=3(x-5)^2-2`

Ta thấy : `3(x-5)^2>=0`

`->3(x-5)^2-2>=-2`

Dấu = xảy ra `<=>x-5=0`          `<=>x=5`

Vậy GTNN của biểu thức là `-2` khi `x=5`

`2x^2+12x+19`

`=2(x^2+6x+19/2)`

`=2(x^2+2.x.3+9+1/2)`

`=2(x^2+2.x.3+3^2)+2.1 /2`

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

Ta thấy : `2(x+3)^2>=0`

`=>2(x+3)^2+1>=1>0`

Vậy đa thức đã cho vô nghiệm