\(a-3\sqrt{ab}+5b\) với \(a...">
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7 tháng 10 2021

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

\(=3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)=\left(3\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

b) \(5a+3\sqrt{ab}-8b=5a-5\sqrt{ab}+8\sqrt{ab}-8b\)

\(=5\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+8\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(5\sqrt{a}+8\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)

7 tháng 10 2021

a) (\(\sqrt{a}-\sqrt{b}\))(3\(\sqrt{a}+b\))

b) \(\left(\sqrt{a}-\sqrt{b}\right)\left(5\sqrt{a}+8\sqrt{b}\right)\)

 

14 tháng 7 2018

a/ \(A=xy+y\sqrt{x}+\sqrt{x}+1\left(x\ge0\right)\)

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

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

b/ \(B=x-3\sqrt{xy}+2y\left(x\ge0;y\ge0\right)\)

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

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

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

c/\(C=2a-7\sqrt{ab}+5b\left(x\ge0;y\ge0\right)\)

\(=2a-2\sqrt{ab}-5\sqrt{ab}+5b\)

\(=2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-5\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}-5\sqrt{b}\right)\)

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: \(x,y\ge0\)

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

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

14 tháng 8 2019

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

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

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

14 tháng 8 2019

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

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

21 tháng 6 2017

đk : \(a\ge0;b\ge0;a\ne b\)

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

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

b) đk : \(a\ge0;b\ge0;a\ne b\)

\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

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

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

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