\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{5x+10+14x-28-20}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19}{2\left(x+2\right)}\\ c,x=-\dfrac{1}{2}\Leftrightarrow A=\dfrac{19}{2\left(2-\dfrac{1}{2}\right)}=\dfrac{19}{2\cdot\dfrac{3}{2}}=\dfrac{19}{3}\)

=4x^2-4x+1+x^2+6x+9-5x^2+245

=2x+255

Lời giải:

ĐKXĐ: $x\geq 0$

$=\frac{(\sqrt{x}-2)(\sqrt{x}+2)+10-x}{\sqrt{x}+2}=\frac{x-4+10-x}{\sqrt{x}+2}=\frac{6}{\sqrt{x}+2}$

\(A=\left|x-7\right|+2x-3=x-7+2x-3=3x-10\)

\(\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

\(=\left(x-1\right)^2=x^2-2x+1\)

Chọn A

A

chon A bạn nhé

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{x-3\sqrt{x}+8}{x-7\sqrt{x}+10}-\dfrac{\sqrt{x}-1}{\sqrt{x}-5}\left(x\ge0,x\ne\left\{4;25\right\}\right)\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{x-3\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}-5}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)+x-3\sqrt{x}+8-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{x-5\sqrt{x}+x-3\sqrt{x}+8-\left(x-3\sqrt{x}+2\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{x-5\sqrt{x}+6}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}-5}\)

Để A nguyên : \(\dfrac{\sqrt{x}-3}{\sqrt{x}-5}=1+\dfrac{2}{\sqrt{x}-5}\in Z\)

\(=>\dfrac{2}{\sqrt{x}-5}\in Z=>\sqrt{x}-5\in\left\{1;-1;2;-2\right\}\)

\(=>\sqrt{x}\in\left\{6;4;7;3\right\}\\ =>x\in\left\{36;16;49;9\right\}\) (TMDK)

x ϵ {36;16;49;9}