Question

Bài 1. tìm điều kiện xác định và tính giá trị các biểu thức sau :

1) A= \(\frac{\sqrt{x}-1}{\sqrt{x}+1}khi\) x =\(4-2\sqrt{3}\)

2) B= \(\frac{\sqrt{x}}{\sqrt{x}-\sqrt{2}}\) khi x =\(5+2\sqrt{6}\)

Bài 2. Tìm điều kiện xác định và rút gọn các biểu thức sau :

1) A= \(\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}\)

2) B = \(\frac{3\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{10\sqrt{x}}{x-4}\)

3) C = \(\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}-\frac{5\sqrt{x}+5}{x-4}\right).\frac{x-4}{\sqrt{x}}\)

Answer 1

Bài 1

1)

Đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

Khi đó A=\(\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)

2) Đề là \(5-2\sqrt{6}\)sẽ hợp lý hơn nha bn

Đkxđ\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-\sqrt{2}\ne0\end{matrix}\right.\)

Ta có \(5-2\sqrt{6}=\left(1-\sqrt{6}\right)^2\)

Khi đó

B= \(\frac{1-\sqrt{6}}{1-\sqrt{6}-\sqrt{2}}\)

Answer 2

1)

đk: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Rgọn

A=\(\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}\)

= \(\frac{x+12+\sqrt{x}-2-\left(4\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

2)

B=\(\frac{3\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{10\sqrt{x}}{x-4}\) đk \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

= \(\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

= \(\frac{3x-5\sqrt{x}-2-\left(x+3\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{3x-5\sqrt{x}-2-x-3\sqrt{x}-2+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{2x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{\left(2x+2\sqrt{x}\right)-\left(4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{2\sqrt{x}\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{\left(\sqrt{x}+2\right)2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)

Chúc bn học tốt

Nhớ tích cho mk nhé