Tìm đkxđ và rút gọn biểu thức:
C= $(\frac{\sqrt[]{x}-2}{x-1}-$ $\frac{\sqrt[]{x}+2}{x+2\sqrt[]{x}+1})$$.\frac{(1-x)^2}{2}$
Tìm đkxđ và rút gọn biểu thức:
C= $(\frac{\sqrt[]{x}-2}{x-1}-$ $\frac{\sqrt[]{x}+2}{x+2\sqrt[]{x}+1})$$.\frac{(1-x)^2}{2}$
Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} x – 1 \ne 0\\ x \ge 0\\ x + 2\sqrt x + 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne 1\\ x \ge 0\\ {\left( {\sqrt x + 1} \right)^2} \ne 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} x \ne 1\\ x \ge 0 \end{array} \right. \end{array}$
$\begin{array}{l} C\left( {\dfrac{{\sqrt x – 2}}{{x – 1}} – \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\ C = \left( {\dfrac{{\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}} – \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\ C = \dfrac{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 1} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}{{\left( {x – 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\ C = \dfrac{{x – \sqrt x – 2 – x – \sqrt x + 2}}{{\left( {x – 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\ C = \dfrac{{ – \sqrt x \left( {x – 1} \right)}}{{\sqrt x + 1}}\\C=\dfrac{-\sqrt x(\sqrt x-1)(\sqrt x+1)}{\sqrt x +1}\\C=-\sqrt{x}(\sqrt x-1) \end{array}$
Điều kiện xác định:`x>=0,x ne 1`
`C=((sqrtx-2)/(x-1)-(sqrtx+2)/(x+2sqrtx+1)).(1-x)^2/2`
`=(((sqrtx-2)(sqrtx+1)-(sqrtx+2)(sqrtx-1))/((x-1)(sqrtx+1))).(x-1)^2/2`
`=((x-sqrtx-2-x-sqrtx+2)/((x-1)(sqrtx+1))).(x-1)^2/2`
`=(-(2sqrtx)/((x-1)(sqrtx+1))).(x-1)^2/2`
`=(-(sqrtx(x-1)^2))/((x-1)(sqrtx+1))`
`=-(sqrtx)(sqrtx-1)`
`=sqrtx-x`.