Question

Cho biểu thức :

A= \(\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{x^2-2x+1}{2}\right)\)

a) Xác định x để A tồn tại .

b) Rút gọn .

c) Tìm x thuộc Z để A nhận giá trị nguyên .

d) Tìm x để A nhận giá trị âm .

Answer 1

a) A xác định khi\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) b) Rút gọn: \(A=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{x^2-2x+1}{2}\right)=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)^2+\left(-\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(x-1\right)^2}{2}=\dfrac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)\left(x-1\right)}{2}=\dfrac{x\sqrt{x}-x-4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2}=\dfrac{\left(x\sqrt{x}-x-4\sqrt{x}\right)\left(\sqrt{x}-1\right)}{2}=\dfrac{x^2-x\sqrt{x}-x\sqrt{x}+x-4x+4\sqrt{x}}{2}=\dfrac{x^2-3x-2x\sqrt{x}+4\sqrt{x}}{2}\)chắc sai r nha bạn