`x^{2}+y^{2}-a^{2}-b^{2}-2xy+2ab`

`=(x^{2}-2xy+y^{2})-(a^{2}-2ab+b^{2})`

`=(x-y)^{2}-(a-b)^{2}`

`=[(x-y)+(a-b)].[(x-y)-(a-b)]`

`=(x-y+a-b)(x-y-a+b)`

`81x(x-1)^{2}-4x^{3}`

`=x[81(x-1)^{2}-4x^{2}]`

`=x{[9(x-1)]^{2}-(2x)^{2}}`

`=x[(9x-9)^{2}-(2x)^{2}]`

`=x(9x-9-2x)(9x-9+2x)`

`=x(7x-9)(11x-9)`

`(a):`

Với `x\in ZZ=>x+7;2-x\in ZZ`

Do đó nên để `A` là số hữu tỉ thì : `2-x\ne0 =>x\ne 2`

Vậy để `A` là số hữu tỉ thì : `x\inZZ;x\ne2`

`(b):`

`A` là số hữu tỉ không dương `=>A\le 0`

`=>(x+7)/(2-x)\le 0=>(x+7)/(x-2)\ge 0`

`=>` \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+7\ge0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+7\le0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-7\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-7\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}x>2\\x\le-7\end{matrix}\right.\)

Kết hợp ĐK `A` là số hữu tỉ : `x\in ZZ;x\ne 2`

`=>` Để `A` là số hữu tỉ không dương thì : `x\in ZZ;x\ne {2;1;0;-1;-2;-3;-4;-5;-6}`

`(a):A=(((\sqrt{a}+1)(a-\sqrt{a}+1))/(\sqrt{a}+1)-\sqrt{a}).((\sqrt{a}+1)^{2})/((a-1)^{2})+\sqrt{a}\     (a>0;a\ne 1)`

`=(a-\sqrt{a}+1-\sqrt{a}).((\sqrt{a}+1)^{2})/((\sqrt{a}-1)^{2}.(\sqrt{a}+1)^{2})+\sqrt{a}`

`=(\sqrt{a}-1)^{2}.(1)/((\sqrt{a}-1)^{2})+\sqrt{a}`

`=\sqrt{a}+1`

`(b):A=3<=>\sqrt{a}+1=3`

`<=>\sqrt{a}=2`

`<=>a=4(TMDK)`

Vậy để `A=3` thì `a=4`

\(\left(g\right):x^2+\dfrac{1}{9}=\dfrac{5}{3}:3=\dfrac{5}{3}.\dfrac{1}{3}=\dfrac{5}{9}\\ =>x^2=\dfrac{5}{9}-\dfrac{1}{9}=\dfrac{4}{9}=\left(\pm\dfrac{2}{3}\right)^2\\ =>x=\pm\dfrac{2}{3}\)

\(\sqrt{3x-6}-x=-2\\ < =>\sqrt{3x-6}=x-2\\ < =>\left\{{}\begin{matrix}x-2\ge0\\3x-6=\left(x-2\right)^2=x^2-4x+4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\ge2\\x^2-7x+10=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\ge2\\\left(x-5\right)\left(x-2\right)=0\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\end{matrix}\right.< =>x\in\left\{5;2\right\}=>S=\left\{5;2\right\}\)

`(1):`

Ta xét : `49x^{2}-24x+4=(7x)^{2}-2.7x.(12)/(7)+((12)/(7))^{2}-((12)/(7))^{2}+4`

`=(7x-(12)/(7))^{2}+(52)/(49)\ge (52)/(49)>0` \(\forall x\in R\)

Vậy biểu thức `\sqrt{49x^{2}-24x+4}` luôn xác định với mọi `x\in RR`

`(2):`

Để biểu thức xác định thì :

`x^{2}-9\ge 0<=>x^{2}\ge 9`

`<=>|x|\ge 3`

`=>x\ge 3` hoặc `x\le -3`

Vậy biểu thức `\sqrt{ x^{2}-9}` xác định khi `x\ge 3` hoặc `x\le -3`

\(\sqrt{x^2-6x+9}-3x=5\\ < =>\sqrt{\left(x-3\right)^2}=3x+5\\ < =>\left|x-3\right|=3x+5\left(ĐK:3x+5\ge0< =>x\ge-\dfrac{5}{3}\right)\\ =>\left[{}\begin{matrix}x-3=3x+5\\x-3=-\left(3x+5\right)\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x-x=-5-3\\3x+x=3-5\end{matrix}\right.\\ =>\left[{}\begin{matrix}2x=-8\\4x=-2\end{matrix}\right.< =>\left[{}\begin{matrix}x=-4\left(KTMDK\right)\\x=-\dfrac{1}{2}\left(TMDK\right)\end{matrix}\right.\\ =>S=\left\{-\dfrac{1}{2}\right\}\)

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\left(x>0;x\ne1\right)\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-\left(2-x\right)}{x\left(\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x\left(\sqrt{x}+1\right)}{2\sqrt{x}+2-2+x}\\ =\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x\left(\sqrt{x}+1\right)}{x+2\sqrt{x}}\\ =\dfrac{x}{\sqrt{x}-1}\)

\(\left(g\right):-\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ =-\dfrac{5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}\\ =-\dfrac{5}{7}.\dfrac{11}{11}+1+\dfrac{5}{7}\\ =-\dfrac{5}{7}.1+\dfrac{5}{7}+1\\ =0+1=1\\ \left(i\right):\dfrac{-7}{31}+\left(\dfrac{24}{17}+\dfrac{7}{31}\right)=-\dfrac{7}{31}+\dfrac{7}{31}+\dfrac{24}{17}=\dfrac{24}{17}\)