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manh
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HT.Phong (9A5)
8 tháng 10 2023 lúc 7:07

a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)

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

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

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))

\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

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

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

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HT.Phong (9A5)
8 tháng 10 2023 lúc 7:16

c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)

\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

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

\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)

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ngọc ánh 2k8
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Gia Huy
8 tháng 7 2023 lúc 7:23

a.

Với \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) có:

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

b.

Với  \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\) có:

\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{1}{x-4}\right)\\ =(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}-\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}).\left(\dfrac{x-4}{1}\right)\\ =(\dfrac{x-2\sqrt{x}}{x-4}-\dfrac{x+2\sqrt{x}}{x-4}).\left(x-4\right)\\ =\left(\dfrac{x-2\sqrt{x}-x-2\sqrt{x}}{x-4}\right)\left(x-4\right)\\ =\dfrac{-4\sqrt{x}\left(x-4\right)}{x-4}\\ =-4\sqrt{x}=VP\)

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bbiooo
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Nguyễn Lê Phước Thịnh
10 tháng 1 2021 lúc 22:26

a) Ta có: \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)

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

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

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

b) Ta có: \(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)

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

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

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

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

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

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

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Trang Nguyễn
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An Thy
30 tháng 6 2021 lúc 9:25

\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0,x\ne9\right)\)

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

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

\(=\dfrac{x+3\sqrt{x}}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}:\dfrac{7-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{7-\sqrt{x}}=\dfrac{x}{\sqrt{x}-7}\)

\(B=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\left(x>0,x\ne1\right)\)

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

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

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

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pansak9
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Nguyễn Lê Phước Thịnh
13 tháng 9 2023 lúc 20:17

ĐKXĐ: x>=0; x<>1

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

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

\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)

Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:

\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)

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

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minh ngọc
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HT.Phong (9A5)
17 tháng 8 2023 lúc 9:58

\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=-\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)

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HT.Phong (9A5)
17 tháng 8 2023 lúc 10:10

\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\) (ĐK: \(x\ge0\))

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

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

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

______________

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: \(x\ge0;x\ne9\))

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(=\dfrac{3}{\sqrt{x}+3}\)

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minh ngọc
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HT.Phong (9A5)
31 tháng 8 2023 lúc 13:24

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))

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

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

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

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

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

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

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Thanh Trúc
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Nguyễn Việt Lâm
27 tháng 4 2021 lúc 17:26

Để được trợ giúp 1 cách chính xác, bạn cần gõ đề 1 cách đầy đủ (bao gồm dấu ngoặc)

Nhìn đề của bạn thì chắc chắn là thiếu vài dấu ngoặc rồi

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Trang Nguyễn
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Yeutoanhoc
28 tháng 6 2021 lúc 10:32

`a)(sqrtx-3)/(sqrtx-2)-(2sqrtx-1)/(sqrtx-1)+(x-2)/(x-3sqrtx+2)`

`=(x-4sqrtx+3-(2sqrtx-1)(sqrtx-2)+x-2)/(x-3sqrtx+2)`

`=(2x-4sqrtx+1-2x+5sqrtx-2)/(x-3sqrtx+2)`

`=(sqrtx-1)/(x-3sqrtx+2)`

`=1/(sqrtx-2)`

`b)((x+2)/(xsqrtx-1)-sqrtx/(x+sqrtx+1)+1/(1-sqrtx)):(sqrtx-1)/2`

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

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

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

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Nguyễn Lê Phước Thịnh
28 tháng 6 2021 lúc 10:42

a) Ta có: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

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

\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

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

b) Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

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

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

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

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Cá Lệ Kiều
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Nguyễn Lê Phước Thịnh
17 tháng 8 2021 lúc 22:42

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

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

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

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

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