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Thanh Vân Đinh Thị
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PHẠM PHƯƠNG DUYÊN
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Nguyễn Linh Chi
25 tháng 8 2020 lúc 4:29

ĐKXĐ: \(x\ge1\); x khác 2; 3

Ta có: 

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

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

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

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

=> \(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{2-\sqrt{x}}{\sqrt{x}}\)

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Cao Thành Long
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Nguyễn Linh Chi
29 tháng 6 2019 lúc 17:04

ĐK : x>0, x khác 1

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

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

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

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

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Jenny Avery
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Nguyễn Nhã Thanh
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Trình
5 tháng 8 2017 lúc 7:11

Điều kiện : \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{1}{1+\sqrt{x}}\right):\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)

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

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

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

\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

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

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

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Hoàng hôn  ( Cool Team )
20 tháng 9 2019 lúc 21:50

A=(x​+x​+yyxy​​):(xy​+yx​+xy​−xy​−xyx+y​)

=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}=x​+yx+xy​+yxy​​:xy​(xy​+y)(xy​−x)x(xy​−x)xy​+y(xy​+y)xy​−(x+y)(xy​+y)(xy​−x)​

=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}=x​+yx+y​:xy2−x2yx2yx2xy​+xy2+y2xy​−y2xy​+x2xy​​

=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}=x​+yx+y​.xy2+x2yxy2−x2y

=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}=x​+yx+y​.xy(x+y)xy(y​−x​)(x​+y​)​

=\sqrt{y}-\sqrt{x}=y​−x
 

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Park Jimin
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Phùng Minh Quân
17 tháng 10 2018 lúc 20:14

\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

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

\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)

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

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

\(b)\) Ta có : \(R< -1\)

\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)

\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)

\(\Leftrightarrow\)\(4\sqrt{x}< 6\)

\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)

\(\Leftrightarrow\)\(x< \frac{9}{4}\)

Chúc bạn học tốt ~ 

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Trần ngô hạ uyên
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Quynh Existn
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Nguyễn Huy Tú
21 tháng 7 2021 lúc 19:00

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phạm ngọc tịnh
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Ngọc Vĩ
24 tháng 7 2016 lúc 10:29

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

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

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

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

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