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5 tháng 7 2019

a) \(=9x-9\sqrt{xy}+4\sqrt{xy}-4y\)

\(=\left(9x-9\sqrt{xy}\right)+\left(4\sqrt{xy}-4y\right)\)

\(=9\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)+4\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(9\sqrt{x}+4\sqrt{y}\right)\)

b)\(=\left(xy+\sqrt{x}.y\right)+\left(\sqrt{x}+1\right)\)

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

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

5 tháng 7 2019

Thank kill cô :))

4 tháng 10 2020

a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)

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

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

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

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

4 tháng 10 2020

b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

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

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

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

a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)

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

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

b: \(B=5x^2-7x\sqrt{y}+2y\)

\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)

\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)

\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)

20 tháng 11 2021

\(a,=\sqrt{xy}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{xy}+1\right)\left(\sqrt{x}-1\right)\\ b,=\sqrt{xy}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{xy}+1\right)\)

11 tháng 10 2020

a= 98 b=35 c=122 và d=129

11 tháng 10 2020

a, \(5+\sqrt{5}=\sqrt{5}\left(\sqrt{5}+1\right)\)

b, \(a-2\sqrt{a}=\sqrt{a}\left(\sqrt{a}-2\right)\)

c, \(x-\sqrt{xy}=\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\)

d, \(x-y-\sqrt{x}-\sqrt{y}\)

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

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

a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)

\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)

\(=-7\sqrt{3}\cdot\sqrt{xy}\)

b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)

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

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

$a)-7xy.\sqrt{\dfrac{3}{xy}}$

$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$

$=-7.\sqrt{\dfrac{xy}{3}}$

$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$

$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$

$=(\sqrt{a}+1)(b\sqrt{a}+1)$

26 tháng 9 2021

a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)

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

a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)

b: \(ab+b\sqrt{a}+\sqrt{a}+1\)

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

d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

1 tháng 8 2023

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

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

\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

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

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

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

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

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

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

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

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

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

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

\(D=0\)