Chương I - Căn bậc hai. Căn bậc ba

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Naa.Khahh

giúp em bài 2 câu b ạ!!!

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Yeutoanhoc
27 tháng 6 2021 lúc 16:46

`2)B=(sqrtx+1)/(x-1)-(x+2)/(xsqrtx-1)-(sqrtx+1)/(x+sqrtx+1)(x>0,x ne 1)`

`=(sqrtx+1)/(x-1)-(x+2)/(xsqrtx-1)-(x-1)/(xsqrtx-1)`

`=(sqrtx+1)/(x-1)-(x+2+x-1)/(xsqrtx-1)`

`=(sqrtx+1)/(x-1)-(2x+1)/(xsqrtx-1)`

`=((sqrtx+1)(x+sqrtx+1)-(2x+1)(sqrtx+1))/((x-1)(x-sqrtx+1))`

`=(xsqrtx+2x+2sqrtx+1-2xsqrtx-2x-sqrtx-1)/((x-1)(x-sqrtx+1))`

`=(-xsqrtx+sqrtx)/((x-1)(x-sqrtx+1))`

`=(-sqrtx(x-1))/((x-1)(x-sqrtx+1))`

`=-sqrtx/(x-sqrtx+1)`

Nguyễn Lê Phước Thịnh
27 tháng 6 2021 lúc 18:37

Cách khác:

2) Ta có: \(B=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

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

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

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

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

\(=-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

An Thy
27 tháng 6 2021 lúc 16:43

2b) \(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

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

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

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

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

\(=\dfrac{-\sqrt{x}}{x+\sqrt{x}+1}\)


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