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a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)
\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)-\left(\sqrt{a}-\sqrt{b}\right)\sqrt{ab}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b-\sqrt{ab}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+b\right)\)
b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)
\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)
c)
$\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}$
$=(\sqrt{a^3}+\sqrt{a^2b})-(\sqrt{b^3}+\sqrt{ab^2})$
$=\sqrt{a^2}(\sqrt{a}+\sqrt{b})-\sqrt{b^2}(\sqrt{b}+\sqrt{a})$
$=a(\sqrt{a}+\sqrt{b})-b(\sqrt{b}+\sqrt{a})$
$=(\sqrt{a}+\sqrt{b})(a-b)=(\sqrt{a}+\sqrt{b})^2(\sqrt{a}-\sqrt{b})$
d)
$x-y+\sqrt{xy^2}-\sqrt{y^3}$
$=(x-y)+(\sqrt{xy^2}-\sqrt{y^3})$
$=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})+y(\sqrt{x}-\sqrt{y})$
$=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}+y)$
a)
$3-\sqrt{3}+\sqrt{15}-3\sqrt{15}$
$=\sqrt{3}(\sqrt{3}-1)-\sqrt{15}(3-1)$
$=(\sqrt{3}(\sqrt{3}-1)-\sqrt{15}(\sqrt{3}+1)(\sqrt{3}-1)$
$=(\sqrt{3}-1)[\sqrt{3}-\sqrt{15}(\sqrt{3}+1)]$
$=(\sqrt{3}-1)(\sqrt{3}-\sqrt{45}-\sqrt{15})$
b)
$\sqrt{1-a}+\sqrt{1-a^2}=\sqrt{1-a}+\sqrt{(1-a)(1+a)}$
$=\sqrt{1-a}+\sqrt{1-a}.\sqrt{1+a}=\sqrt{1-a}(1+\sqrt{1+a})$
a/
Bạn coi lại đề, số cuối là \(3\sqrt{15}\) hay \(3\sqrt{5}\)
b/
\(=\sqrt{1-a}+\sqrt{\left(1-a\right)\left(1+a\right)}=\sqrt{1-a}\left(1+\sqrt{1+a}\right)\)
c/
\(=\sqrt{a^3}+\sqrt{a^2b}-\sqrt{b^3}-\sqrt{ab^2}\)
\(=\sqrt{a^2}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b^2}\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)\)
(Hoặc có thể biến đổi thêm \(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}-\sqrt{b}\right)\) cũng được)
d/
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{y^2}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)
a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)
\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)
\(=-7\sqrt{3}\cdot\sqrt{xy}\)
b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
$a)-7xy.\sqrt{\dfrac{3}{xy}}$
$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$
$=-7.\sqrt{\dfrac{xy}{3}}$
$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$
$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$
$=(\sqrt{a}+1)(b\sqrt{a}+1)$
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: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
a)\(3-\sqrt{3}+\sqrt{15}-3\sqrt{5}=\sqrt{3}\left(\sqrt{3}-1\right)-\sqrt{15}\left(\sqrt{3}-1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{15}\right)=\sqrt{3}\left(\sqrt{3}-1\right)\left(1-\sqrt{5}\right)\)\(\)b)\(\sqrt{1-a}+\sqrt{1-a^2}=\sqrt{1-a}.1+\sqrt{1-a}.\sqrt{1+a}=\sqrt{1-a}\left(\sqrt{1+a}+1\right)\)
c)\(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b+\sqrt{ab}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(a+2\sqrt{ab}+b\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2\)