Question

\(A=\left(\frac{3x+\sqrt{9x}-3}{x-\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-2}\right):\frac{1}{x-1}\) Với x >=0,x khác 1

a) Rút gọn A

b) Tìm số tự nhiên x đề 1/A là số tự nhiên

 

Answer 1

thế x=4 thì sao

Answer 2

\(dk:x\ne\left\{1,\sqrt{2},4\right\};x\ge0\)dat \(\sqrt{x}=t\)

\(A=\left(\frac{3t^2}{t^2-t-2}+\frac{1}{t-1}+\frac{1}{t-2}\right)\left(t^2-1\right)==\left(\frac{3t^2}{\left(t-2\right)\left(t-1\right)}+\frac{1}{t-1}+\frac{1}{t-2}\right)\left(t^2-1\right)\)

\(=\left(\frac{3t^2}{\left(t-2\right)\left(t-1\right)}+\frac{t-2}{t-1}+\frac{t-1}{t-2}\right)\left(t-1\right)\left(t+1\right)=3t^2+2t-3\)

\(A=3x+2\sqrt{x}-3\)

b

\(\frac{1}{A}=\frac{1}{3x+2\sqrt{x}-3}\Rightarrow\orbr{\begin{cases}3x+2\sqrt{x}-3=-1\\3x+2\sqrt{x}-3=1\end{cases}}\)tư làm tiếp

Answer 3

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

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

\(=\frac{3x\sqrt{x}-6\sqrt{x}+1+x-\sqrt{x}-2+x-1}{\left(\sqrt{x}-2\right)\left(x-1\right)}:\frac{1}{x-1}\)

\(=\frac{3x\sqrt{x}+2x-7\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(x-1\right)}:\frac{1}{x-1}\)

\(=\frac{3x\sqrt{x}+2x-7\sqrt{x}-2}{\sqrt{x}-2}\)

b/ \(\frac{1}{A}=\frac{3x\sqrt{x}+2x-7\sqrt{x}-2}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{3x\sqrt{x}+2x-7\sqrt{x}-2}\)

Tìm dược x = 4 đó