\(2^{17-x}=2^{12}.2^4\\ 2^{17-x}=2^{16}\\ 17-x=16\\ x=1\)

\(\text{a},=\left[\dfrac{x+2\sqrt{xy}+y-4\sqrt{xy}-2x+2\sqrt{xy}}{x-\sqrt{xy}}\right].\dfrac{\sqrt{x}\left(x-y\right)}{\sqrt{x}+\sqrt{y}}=\left(\dfrac{-x+y}{x-\sqrt{xy}}\right).\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{1}=\dfrac{-x+y}{x-\sqrt{xy}}.\dfrac{x-\sqrt{xy}}{1}=-x+y\\ b=\left(\dfrac{\sqrt{\text{a}^3}-\sqrt{m^3}}{\sqrt{\text{a}}-\sqrt{m}}+\sqrt{\text{a}m}\right).\left(\dfrac{1}{\sqrt{\text{a}}+\sqrt{m}}\right)^2=\left(\text{a}+\sqrt{\text{a}m}+m+\sqrt{\text{a}m}\right).\dfrac{1}{\left(\sqrt{\text{a}}+\sqrt{m}\right)^2}=\dfrac{\left(\sqrt{\text{a}}+\sqrt{m}\right)^2}{\left(\sqrt{\text{a}}+\sqrt{m}\right)^2}=1\)

\(\text{a},=\dfrac{\sqrt{m^3}-\sqrt{y^3}}{\sqrt{m}-\sqrt{y}}-\dfrac{\sqrt{my}\left(\sqrt{m}+\sqrt{y}\right)}{\sqrt{m}+\sqrt{y}}=m+\sqrt{my}+y-\sqrt{my}=m+y\\ b,=\left(\dfrac{\sqrt{b^3}+1}{\sqrt{b}+1}\right).\dfrac{\sqrt{b}-1}{b-1}=\left(b-\sqrt{b}+1\right).\left(\dfrac{1}{\sqrt{b}+1}\right)=\dfrac{b-\sqrt{b}+1}{\sqrt{b}+1}\)

đk a , m > 0 , y> 0

đk b b > 0 

C

\(x^2=64:16\\ x^2=4\)

=> x = 2 ; x=-2

\(\left(xy\right)^2\left(5x+3y\right)^2=x^2y^2\left(25x^2+3xy+9y^2\right)=25x^4y^2+3x^3y^3+9x^2y^4\)

\(\left(\dfrac{1}{3x}+2\text{z}\right)^2=\dfrac{1}{9x^2}+2.\dfrac{1}{3x}.2\text{z}+4\text{z}^2=\dfrac{1}{9x^2}+\dfrac{4}{3}x\text{z}+4\text{z}^2\)

\(\left(2x+\dfrac{1}{4y}\right)^2=4x^2+2.2x.\dfrac{1}{4y}+\dfrac{1}{16y^2}=4x^2+xy+\dfrac{1}{16y^2}\)

đau bai đúng ko vay :0