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\(A,ĐKXĐ:x;y\ge0\)

\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)

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

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

\(ĐKXĐ:x;y\ge0\)

\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)

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

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

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

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

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

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) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)

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

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

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

11 tháng 9 2018

với a,b,x,y không âm ta có

a,\(ab+b\sqrt{a}+\sqrt{a}+1\)

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

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

b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

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

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

24 tháng 4 2017

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

b. \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(x+2\sqrt{xy}+y\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)

5 tháng 7 2015

\(\text{a) }\sqrt{a^3+b^3}+\sqrt{a^2-b^2}=\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}+\sqrt{\left(a+b\right)\left(a-b\right)}\)

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

\(\text{b) }\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{xy}\text{ không phân tích được.}\)

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

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

\(\text{d) }a+5\sqrt{a}+4=\sqrt{a}.\sqrt{a}+\sqrt{a}+4\sqrt{a}+4=\sqrt{a}\left(\sqrt{a}+1\right)+4\left(\sqrt{a}+1\right)\)

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

20 tháng 6 2017

a, \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\)

\(=\dfrac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{7}}\)

b, \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)

\(=\dfrac{2.\sqrt{5}.\sqrt{3}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{3}+\sqrt{2}.\sqrt{3}}{2.\sqrt{5}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}+\sqrt{2}.\sqrt{3}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}.\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}.\left(1-\sqrt{2}\right)-\sqrt{3}.\left(1-\sqrt{2}\right)}\)

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

c, \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}.\sqrt{x}+\sqrt{x}.\sqrt{y}}{\sqrt{y}.\sqrt{y}+\sqrt{x}.\sqrt{y}}\)

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

Chúc bạn học tốt!!!

20 tháng 6 2017

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

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

22 tháng 6 2019

\(ab+b\sqrt{a}+\sqrt{a}+1\)

(đk: \(a\ge0\))

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

22 tháng 6 2019

ĐK: \(x,y\ge0\)

\(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=x\left(\sqrt{x}+\sqrt{y}\right)-y\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)

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

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, \(x-y-\sqrt{x}-\sqrt{y}\)

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

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