Question

cho biểu thức A=\(\frac{X\sqrt{X}+1}{X-1}\)-\(\frac{X-1}{\sqrt{X}+1}\)

a tìm ĐKXĐ và rút gọ A

B. tính giá trị bt A khi x=\(\frac{9}{4}\)

c.tìm tất cả các giá trị của x để A<1

Answer 1

a) ĐKXĐ: \(x\ne1\)

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

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

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

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

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

b) \(A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=3\)

c) \(A< 1\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-1}< 1\)

\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\frac{1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}< 1\)

\(\Leftrightarrow0\le x< 1\)

Vậy....

Answer 2

a, ĐKXĐ : \(\left\{{}\begin{matrix}X-1\ne0\\\sqrt{X}+1\ne0\\X\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}X\ne1\\X\ge0\end{matrix}\right.\)

- Ta có : \(A=\frac{X\sqrt{X}+1}{X-1}-\frac{X-1}{\sqrt{X}+1}\)

=> \(A=\frac{X\sqrt{X}+1}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}-\frac{\left(X-1\right)\left(\sqrt{X}-1\right)}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}\)

=> \(A=\frac{X\sqrt{X}+1-\left(X-1\right)\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)

=> \(A=\frac{X\sqrt{X}+1-X\sqrt{X}+\sqrt{X}+X-1}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)​