($\frac{√x+2}{x+2√x+1}$-$\frac{√x-2}{x-1}$).$\frac{√x-2}{x-1}$ rút gọn hộ mình nha
2 câu trả lời
$(\dfrac{\sqrt[]{x}+2}{x+2\sqrt[]{x}+1}-$ $\dfrac{\sqrt[]{x}-2}{x-1}).$ $\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}}$
$=(\dfrac{\sqrt[]{x}+2}{(\sqrt[]{x}+1)^2}-$ $\dfrac{\sqrt[]{x}-2}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}).$ $\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}}$
$=\dfrac{(\sqrt[]{x}+2)(\sqrt[]{x}-1)-(\sqrt[]{x}-2)(\sqrt[]{x}+1)}{(\sqrt[]{x}+1)^2.(\sqrt[]{x}-1)}.$ $\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}}$
$\dfrac{x-\sqrt[]{x}+2\sqrt[]{x}-2-(x+\sqrt[]{x}-2\sqrt[]{x}-2)}{(\sqrt[]{x}+1)^2.(\sqrt[]{x}-1)}.$ $\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}}$
$=\dfrac{x+\sqrt[]{x}-2-x+\sqrt[]{x}+2}{(\sqrt[]{x}+1)^2.(\sqrt[]{x}-1)}.$ $\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}}$
$=\dfrac{2\sqrt[]{x}}{(\sqrt[]{x}+1)^2.(\sqrt[]{x}-1)}.$ $\dfrac{\sqrt[]{x}+1}{\sqrt[]{x}}$
$=\dfrac{2(\sqrt[]{x}+1)}{(\sqrt[]{x}+1)^2.(\sqrt[]{x}-1)}$
$=\dfrac{2}{(\sqrt[]{x}-1)(\sqrt[]{x}+1)}$
$=\dfrac{2}{x-1}$
Chúc bạn học tốt !!!!
Đáp án + Giải thích các bước giải:
`((\sqrt{x} + 2)/(x+2\sqrt{x} + 1) - (\sqrt{x} - 2)/(x-1) ) . ((\sqrt{x}+1)/(\sqrt{x}))`
đk : `x` $\neq$ `1`
`=( (\sqrt{x} + 2)/(\sqrt{x}+1)^2 - (\sqrt{x}-2)/((\sqrt{x}+1)(\sqrt{x}-1)) ) . ((\sqrt{x}+1)/(\sqrt{x}))`
`=( (\sqrt{x}+2)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+1))/((\sqrt{x}+1)(x-1)) . ((\sqrt{x}+1)/(\sqrt{x}))`
`=(x+\sqrt{x}-2-(x-\sqrt{x}-2))/((\sqrt{x}+1)(x-1)) . ((\sqrt{x}+1)/(\sqrt{x}))`
`=(x+\sqrt{x}-2-x+\sqrt{x}+2)/((\sqrt{x}+1)(x-1)) . ((\sqrt{x}+1)/(\sqrt{x}))`
`=(2\sqrt{x})/((\sqrt{x}+1)(x-1)) . (\sqrt{x}+1)/\sqrt{x} `
`=(2(\sqrt{x}+1))/((\sqrt{x}+1)(x-1)) `