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Đinh Trần Tiến
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Nguyễn Ngọc Thùy Duyên
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Nguyễn Lê Phước Thịnh
17 tháng 7 2021 lúc 11:18

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

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

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

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

 

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Hồng Nhan
11 tháng 6 2021 lúc 21:15

Không có mô tả.

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Hồng Nhan
11 tháng 6 2021 lúc 21:16

Không có mô tả.

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Trang Nguyễn
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Nguyễn Lê Phước Thịnh
10 tháng 7 2021 lúc 10:20

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

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

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

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

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

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nchdtt
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Nguyễn Việt Lâm
22 tháng 7 2021 lúc 7:54

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

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

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

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

b. Đặt \(B=A-2x\)

\(B=\sqrt{x}-1-2x=-2\left(\sqrt{x}-\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}\)

\(B_{max}=-\dfrac{7}{8}\) khi \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

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Trang Nguyễn
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Nguyễn Lê Phước Thịnh
28 tháng 6 2021 lúc 11:34

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

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

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

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

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

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

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

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

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

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

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

 

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Nguyễn Lê Phước Thịnh
28 tháng 6 2021 lúc 11:36

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

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

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

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

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Nguyễn Ngọc Lộc
28 tháng 6 2021 lúc 11:37

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

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

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

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

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

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

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

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

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

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

 

 

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Lê Thu Trang
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Dương Thị Trà My
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Phương An
30 tháng 7 2017 lúc 21:23

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

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

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

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Aikatsu
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