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Nàng tiên cá
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Biển Ác Ma
30 tháng 7 2019 lúc 18:55

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

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

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

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

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

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

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

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

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

wary reus
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Hoàng Lê Bảo Ngọc
4 tháng 9 2016 lúc 10:54

Sai đề

Nguyễn Ngọc Anh
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Nguyễn Nhã Thanh
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Đinh Đức Hùng
11 tháng 8 2017 lúc 16:20

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

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

\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}\)

\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}\)

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

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

Thân Thùy Dương
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\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

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

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

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

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

_Minh ngụy_

\(\(d)\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)\)

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

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

\(\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)\)( vì \(\(x\ge0\)\))

_Minh ngụy_

tiểu long nữ
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Mafia
12 tháng 5 2018 lúc 18:27

\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)

\(A=\left[5-\frac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right]\left[5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right]\)

\(A=\left[5-\sqrt{xy}\right]\left[5+\sqrt{xy}\right]\)

\(A=25-xy\)

vậy \(A=25-xy\)

Despacito
12 tháng 5 2018 lúc 18:30

\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)

\(A=\left(5-\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)

\(A=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)

\(A=25-xy\)

Namikaze Minato
12 tháng 5 2018 lúc 18:37

\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\left(5-\frac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)

\(=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)

\(=25-xy\)

Nguyễn Hương Ly
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Đặng Trúc Anh
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Bùi Thế Hào
7 tháng 8 2017 lúc 12:11

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

=> \(A=\left(2\sqrt{xy}+x+y\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

=> \(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2}=1\)

ĐS: A=1

nguyenthiluyen
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