`3 sqrt(x)-1=40`

`=>3 sqrt(x)=39`

`=>sqrt(x)=13`

`=>(sqrt(x))^2=13^2`

`=>x=13^2=169`

`(x-1)^2=9/16`

`(x-1)^2=(3/4)^2`

`=>` \(\left[{}\begin{matrix}x-1=\dfrac{3}{4}\\x-1=-\dfrac{3}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Có 

`a+5b vdots 7`

`=>4(a+5b) vdots 7`

`=>4a+20b vdots 7`

Ta có :

`(4a+20b)+(10a+b)`

`=4a+20b+10a+b`

`=(4a+10a)+(20b+b)`

`=14a+21b=7(2a+3b) vdots 7`

hay `(4a+20b)+(10a+b) vdots 7`

mà `4a+20b vdots 7(cmt)`

nên `10a+b vdots 7(đpcm)`