Question

[Sonyeondan Bangtan]

1. Tìm điều kiện để biểu thức có nghĩa:

a) A = \(\frac{1}{1-\sqrt{x-1}}\)

b) B = \(\frac{1}{\sqrt{x^2-2x+1}}\)

2. Rút gọn biểu thức:

B = \(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}\)

3. Q = \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

a) Tìm x để Q có nghĩa

b) Rút gọn Q

c) Tìm x \(\in\) Z để Q \(\in\) Z

Answer 1

[@Nk>↑@]

1.a)ĐKXĐ:\(\left\{{}\begin{matrix}x-1\ge0\\1-\sqrt{x-1}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ne2\end{matrix}\right.\)

b)ĐKXĐ:\(\left\{{}\begin{matrix}x^2-2x+1\ge0\\\sqrt{x^2-2x+1}\ne0\end{matrix}\right.\Leftrightarrow x^2-2x+1>0\Leftrightarrow\left(x-1\right)^2>0\) luôn đúng với mọi x \(\ne\)1

Vậy biểu thức xác định khi \(x\ne1\)

2.\(B=\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{60}-2}=\frac{\sqrt{15-2\sqrt{15}+1}}{2\sqrt{15}-2}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)

3.a)ĐKXĐ:\(x\ge0\)

b)\(Q=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(x+1\right)}\right):\frac{x-\sqrt{x}+1}{\sqrt{x}^3+1}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}\left(x+1\right)}.\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

c)\(Q=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)

Để \(Q\in Z\) thì

\(1⋮\sqrt{x}\)

\(\Rightarrow\sqrt{x}\in\left\{-1;1\right\}\)(loại -1 vì \(\sqrt{x}\ge0\))

\(\Rightarrow x\in\left\{1\right\}\)

Answer 2

[你混過 vulnerable 他 難怪歐長]

cho mik hoi \(x\in Z\) \(Q\in Z\) khi nao v