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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: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
\(\text{a) }\sqrt{a^3+b^3}+\sqrt{a^2-b^2}=\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}+\sqrt{\left(a+b\right)\left(a-b\right)}\)
\(=\sqrt{a+b}\left(\sqrt{a^2-ab+b^2}+\sqrt{a-b}\right)\)
\(\text{b) }\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{xy}\text{ không phân tích được.}\)
\(\text{c) }=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\left(\sqrt{x}-\sqrt{y}\right).\sqrt{xy}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+2\sqrt{xy}\right)\)\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(\text{d) }a+5\sqrt{a}+4=\sqrt{a}.\sqrt{a}+\sqrt{a}+4\sqrt{a}+4=\sqrt{a}\left(\sqrt{a}+1\right)+4\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+4\right)\)
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)=a+\(2\sqrt{a}+2\sqrt{a}\)+4
=\(\sqrt{a}\left(\sqrt{a}+2\right)+2\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)^2\)
b)= \(\left(a-\sqrt{7}\right)\left(a+\sqrt{7}\right)\)
c) \(\sqrt{a}\left(\sqrt{b}-4\right)+3\cdot\left(\sqrt{b}-4\right)=\left(\sqrt{a}+3\right)\left(\sqrt{b}-4\right)\)
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: \(B=5x^2-7x\sqrt{y}+2y\)
\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)
\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)
a)\(a-5\sqrt{a}=\sqrt{a}\left(\sqrt{a}-5\right)\)
b)\(a-7=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)
c)\(a+4\sqrt{a}+4=\left(\sqrt{a}+2\right)^2\)
d)\(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12=\sqrt{x}\left(\sqrt{y}-4\right)+3\left(\sqrt{y}-4\right)=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)
em cảm ơn ạ