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Những câu hỏi liên quan
Phạm Trần Phát
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HT.Phong (9A5)
5 tháng 9 2023 lúc 10:39

\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\) (ĐK: \(x>0;x\ne2;x\ne1\))

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

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

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

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

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

Toru
5 tháng 9 2023 lúc 10:44

\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x>0;x\ne1;x\ne4\right)\)

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

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

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

#Urushi

Quynh Existn
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Akai Haruma
17 tháng 7 2021 lúc 22:31

1. ĐKXĐ: $x>0; x\neq 9$

\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)

Akai Haruma
17 tháng 7 2021 lúc 22:38

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

Akai Haruma
17 tháng 7 2021 lúc 22:40

3. ĐKXĐ: $a\geq 0; a\neq 1$

\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)

\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)

 

Nguyễn Châu Mỹ Linh
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Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 13:44

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

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

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

Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 13:46

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

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

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

Phạm Trần Phát
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Nguyễn Lê Phước Thịnh
4 tháng 9 2023 lúc 14:34

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

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

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

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

HT.Phong (9A5)
4 tháng 9 2023 lúc 16:31

\(A=\left(\dfrac{x}{\sqrt{x}-1}-\sqrt{x}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\) (ĐK: \(x>0;x\ne1\))

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

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

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

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

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

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

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

Jackson Williams
4 tháng 9 2023 lúc 15:12

A = \(\dfrac{x}{\sqrt{x}+1}\)

KYAN Gaming
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Nguyễn Lê Phước Thịnh
10 tháng 8 2021 lúc 20:43

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

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

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

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

KYAN Gaming
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An Thy
25 tháng 7 2021 lúc 15:53

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

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

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

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

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

Akai Haruma
25 tháng 7 2021 lúc 15:54

Lời giải:
ĐKXĐ: $x>0; x\neq 1$

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

\(=\left[\frac{1}{\sqrt{x}-1}-\frac{2}{(\sqrt{x}-1)(x+1)}\right].\frac{x+1}{x-\sqrt{x}+1}=\frac{x+1-2}{(\sqrt{x}-1)(x+1)}.\frac{x+1}{x-\sqrt{x}+1}=\frac{x-1}{(\sqrt{x}-1)(x-\sqrt{x}+1)}=\frac{\sqrt{x}+1}{x-\sqrt{x}+1}\)

Nguyễn Lê Phước Thịnh
25 tháng 7 2021 lúc 19:50

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

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

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

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

Nguyễn Thị Thu Phương
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Akai Haruma
6 tháng 8 2021 lúc 17:03

1.

\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)

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

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

Akai Haruma
6 tháng 8 2021 lúc 17:06

2.

\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)

\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)

Akai Haruma
6 tháng 8 2021 lúc 17:09

3.

\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)

\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)

Quynh Existn
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Nguyễn Huy Tú
17 tháng 7 2021 lúc 22:33

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Quynh Existn
17 tháng 7 2021 lúc 22:20

Làm ơn giúp mình với... :(

Quynh Existn
17 tháng 7 2021 lúc 22:38

bạn chụp rõ hơn dc không ạ? :( 

KYAN Gaming
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Trần Ái Linh
21 tháng 7 2021 lúc 21:12

ĐK: `x>0` 

`A=((\sqrtx)/(\sqrtx+2) - 4/(x+2\sqrtx)):(1+1/(\sqrtx))`

`=((\sqrtx)/(\sqrtx+2)-4/(\sqrtx(\sqrtx+2))):((\sqrtx+1)/(\sqrtx))`

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

`=((\sqrtx+2)(\sqrtx-2))/(\sqrtx+2) . 1/(\sqrtx+1)`

`=(\sqrt-2)/(\sqrtx+1)`

Edogawa Conan
21 tháng 7 2021 lúc 21:14

Ta có:\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\)

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

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