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Những câu hỏi liên quan
Duong Thi Nhuong
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Phùng Khánh Linh
30 tháng 6 2018 lúc 16:41

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

\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)

\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)

Thành Trương
1 tháng 7 2018 lúc 9:24

Câu a

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

Thành Trương
1 tháng 7 2018 lúc 9:32

Hệ hai phương trình bậc nhất hai ẩn

ho quoc khanh
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Qúy Công Tử
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Briona
4 tháng 8 2018 lúc 21:44

1. \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

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

\(=\sqrt{a}+2-\sqrt{a}-2\)

= 0

Nguyễn Lê Phước Thịnh
9 tháng 8 2022 lúc 13:46

2: \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\)

\(=\sqrt{x}-\sqrt{y}+\sqrt{y}-\sqrt{x}=0\)

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

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

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

Bạn cần làm gì với biểu thức này?
 

Vu Huyen Anh
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KAl(SO4)2·12H2O
7 tháng 8 2018 lúc 8:34

a) \(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)

\(A=\sqrt{\left(2+\sqrt{3}\right)\left(\sqrt{2+\sqrt{3}}+2\right)\left(-\sqrt{2+\sqrt{3}}+2\right)}\)

\(A=\sqrt{1}\)

\(A=1\)

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

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

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

\(B=\frac{-x^{\frac{5}{2}}\sqrt{y}+\sqrt{x}.y^{\frac{5}{2}}}{\left(\sqrt{xy}-y\right)\left(\sqrt{xy}-x\right)}\)

\(B=\frac{\left(\sqrt{x}.y^{\frac{5}{2}}-x^{\frac{5}{2}}\sqrt{y}\right)\left(y+\sqrt{xy}\right)\left(x+\sqrt{xy}\right)}{\left(-y^2+xy\right)\left(-x^2+xy\right)}\)

c) \(C=\sqrt{\left(3-\sqrt{5}\right)^2+\sqrt{6}-2\sqrt{5}}\)

\(C=14-6\sqrt{5}+\sqrt{6}-2\sqrt{5}\)

\(C=14-8\sqrt{5}+\sqrt{6}\)

\(C=\sqrt{14-8\sqrt{5}+\sqrt{6}}\)

prayforme
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Phương An
7 tháng 8 2017 lúc 15:34

\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)

\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}\)

\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)

= - 1

\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)

\(=\dfrac{\sqrt{5}+1}{2}\)

\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)

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

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

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

= 2

\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)

\(=4+9+16+49\)

= 78

Phương An
7 tháng 8 2017 lúc 15:50

\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)

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

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

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

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

\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)

\(=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

\(=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}\)

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

= 3

Ta có:

\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)

\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)

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

\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)

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

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

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

Nhi Trần
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Seu Vuon
26 tháng 4 2015 lúc 15:24

3)\(...=\left[\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\right].\frac{1-xy}{x+xy}\)

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

phamthiminhanh
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Vũ Minh Phong
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Nguyễn Lê Phước Thịnh
17 tháng 7 2021 lúc 23:32

a) Ta có: \(\sqrt{27\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=3\sqrt{3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)

\(=3\sqrt{15}-9\)

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