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Lê Thị Ngọc Duyên
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Nguyễn Thanh Hải
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Nguyễn Ngọc Huy Toàn
24 tháng 5 2022 lúc 21:13

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

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

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

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

\(:\left(\dfrac{\text{​​}\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)

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

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

\(A=1\)

 

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이성경
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Nguyễn Lê Phước Thịnh
12 tháng 10 2022 lúc 16:04

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

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

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

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

c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

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Minh Anh Vũ
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Trang Nguyễn
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Nguyễn Ngọc Lộc
28 tháng 6 2021 lúc 10:59

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

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

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

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

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

 

 

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Nguyễn Lê Phước Thịnh
28 tháng 6 2021 lúc 11:04

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

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

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

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

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

d) Ta có: \(D=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)

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

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

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

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

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

\(=-1\)

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phamthiminhanh
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Tuyết Linh Linh
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Akai Haruma
2 tháng 3 2021 lúc 20:22

Lời giải:

a) ĐK: $x\geq 0; y\geq 0; x\neq y$

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

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

\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$

 

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Trang Nguyễn
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Nguyễn Lê Phước Thịnh
5 tháng 7 2021 lúc 21:30

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

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

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

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

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Lê Thị Thục Hiền
5 tháng 7 2021 lúc 21:35

Đk:\(xy\ne1;x\ge0;y\ge0\)

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

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

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

b) Áp dụng AM-GM có:

\(1+x\ge2\sqrt{x}\Leftrightarrow\)\(\dfrac{2\sqrt{x}}{1+x}\le1\)

Dấu "=" xảy ra khi x=1 (tm)

Vậy \(P_{max}=1\)

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Ngô Thị Lan Anh
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Akai Haruma
3 tháng 8 2021 lúc 18:04

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