Question

cho A=(\sqrt(x))/(\sqrt(x-2))+(x-3\sqrt(x+8))/(x-7\sqrt(x+10))-(\sqrt(x-1))/(\sqrt(x-5))

rút gọn A

tìm giá trị nguyên của x để biểu thức A nguyên

Answer 1

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

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

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)

Answer 2

x ϵ {36;16;49;9}