Question

[30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ]

  

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(C=\left(\dfrac{x}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

\(=\left(\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\)

\(=\dfrac{x-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{x-4+10-x}\)

\(=\dfrac{x-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}\cdot\dfrac{1}{6}\)

\(=\dfrac{-6}{6\left(\sqrt{x}-2\right)}=\dfrac{-1}{\sqrt{x}-2}\)

b: Khi \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) thì

\(C=\dfrac{-1}{\sqrt{\left(\sqrt{3}-1\right)^2}-2}=\dfrac{-1}{\sqrt{3}-1-2}\)

\(=\dfrac{-1}{\sqrt{3}-3}=\dfrac{1}{3-\sqrt{3}}=\dfrac{3+\sqrt{3}}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(=\dfrac{3+\sqrt{3}}{9-3}=\dfrac{3+\sqrt{3}}{6}\)

c: Để C là số nguyên thì \(-1⋮\sqrt{x}-2\)

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

=>\(\sqrt{x}\in\left\{3;1\right\}\)

=>\(x\in\left\{1;9\right\}\)

d: Để C=5 thì \(-\dfrac{1}{\sqrt{x}-2}=5\)

=>\(\sqrt{x}-2=-\dfrac{1}{5}\)

=>\(\sqrt{x}=\dfrac{9}{5}\)

=>\(x=\left(\dfrac{9}{5}\right)^2=\dfrac{81}{25}\left(nhận\right)\)

e: Để C>=0 thì \(-\dfrac{1}{\sqrt{x}-2}>=0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

Kết hợp ĐKXĐ, ta được: 0<=x<4

Answer 2

[Bùi Thúy Ngọc]

.