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Nguyễn Thị Thu Phương
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Trên con đường thành côn...
9 tháng 8 2021 lúc 20:05

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Phạm Trần Phát
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Nguyễn Lê Phước Thịnh
5 tháng 9 2023 lúc 19:57

21: ĐKXĐ: x>0; x<>1

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

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

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

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

22:
DKXĐ: x>0; x<>1

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

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

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

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

23: ĐKXĐ: x>0; x<>4

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

\(=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\)

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

24: ĐKXĐ: x>=0; x<>1

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

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

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

25:

ĐKXĐ: x>=0; x<>1

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

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

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

27: ĐKXĐ: x>0; x<>4

\(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x-8\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-1-2\sqrt{x}+1}\)

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

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

Nguyễn Thị Thu Phương
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Nguyễn Thị Thu Phương
15 tháng 8 2021 lúc 15:55

ai giúp với ạ :<

Nguyễn Lê Phước Thịnh
15 tháng 8 2021 lúc 20:30

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

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

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

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

Nguyễn Lê Phước Thịnh
15 tháng 8 2021 lúc 23:20

1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{5\sqrt{x}-15}{3x-59}\)

Nguyễn Thị Thu Phương
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Nguyễn Lê Phước Thịnh
15 tháng 8 2021 lúc 23:20

1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{5\sqrt{x}-15}{3x-59}\)

Nguyễn Lê Phước Thịnh
15 tháng 8 2021 lúc 23:22

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

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

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

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

Nguyễn Lê Phước Thịnh
15 tháng 8 2021 lúc 23:26

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

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

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

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

hoàng
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hoàng
15 tháng 9 2023 lúc 15:39

help

Nguyễn Lê Phước Thịnh
15 tháng 9 2023 lúc 18:21

loading...  => đề sai rồi bạn

Nguyễn Thị Thu Phương
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Nguyễn Lê Phước Thịnh
23 tháng 8 2021 lúc 21:13

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

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

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

Nguyễn Lê Phước Thịnh
23 tháng 8 2021 lúc 22:30

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

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

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

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

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

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

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

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

Nguyễn Thị Thu Phương
Xem chi tiết
Nguyễn Thị Thu Phương
11 tháng 8 2021 lúc 20:13

nãy đăng ảnh nhưng không hiện, lại phải mất công đánh lại :Đ

 

Nguyễn Lê Phước Thịnh
11 tháng 8 2021 lúc 20:27

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

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

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

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

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

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

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

 

Nguyễn Lê Phước Thịnh
11 tháng 8 2021 lúc 22:37

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

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

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

Phạm Trần Phát
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HT.Phong (9A5)
4 tháng 9 2023 lúc 10:38

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

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

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

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

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

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

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

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

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

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

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

Jackson Williams
4 tháng 9 2023 lúc 10:42

A = \(\dfrac{1}{\sqrt{x}-x}\) 

quang
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Akai Haruma
15 tháng 4 2023 lúc 22:00

Đoạn $x\sqrt{x}-a$ là sao vậy bạn? Có nhầm lẫn gì không?

Nguyễn Lê Phước Thịnh
15 tháng 4 2023 lúc 23:19

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

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

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