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\(x\sqrt{y}-y\sqrt{x}=\sqrt{x^2}.\sqrt{y}-\sqrt{y^2}.\sqrt{x}=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(\left(x^2+4x+6\right)\left(x^2+6x+6\right)-3x^2\left(1\right)\)
Đặt \(x^2+5x+6=t\)Thay vào (1) ta được:
\(\left(t-x\right)\left(t+x\right)-3x^2\)
\(=t^2-x^2-3x^2\)
\(=t^2-4x^2\)
\(=\left(t-2x\right)\left(t+2x\right)\)Thay \(t=x^2+5x+6\)ta được:
\(\left(x^2+5x+6-2x\right)\left(x^2+5x+6+2x\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+x+6x+6\right)\)
\(=\left(x^2+3x+6\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x^2+3x+6\right)\left(x+1\right)\left(x+6\right)\)
\(=\left(x\sqrt{y}-y\sqrt{x}\right)+\left(x-y\right)\)
\(=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}+\sqrt{x}+\sqrt{y}\right)\)
\(x-6\sqrt{x-3}+6\text{=}x-3-6\sqrt{x-3}+9\)
\(\text{=}\left(\sqrt{x-3}\right)^2-2.3.\sqrt{x-3}+\left(3\right)^2\)
\(\text{=}\left(\sqrt{x-3}-3\right)^2\)
A = \(x-6\)\(\sqrt{x-3}\) + 6 (đkxd \(x>3\))
A = (\(x\) - 3) - 2.3.\(\sqrt{x-3}\) + 9
A = (\(\sqrt{x-3}\))2 - 2.3.\(\sqrt{x-3}\) + 32
A = (\(\sqrt{x-3}\)- 3)2
a,3x-6\(\sqrt{x}\)-6
=2(x-3\(\sqrt{x}\)-3)
=2(x-2\(\sqrt{x}\).\(\frac{3}{2}\)+\(\frac{9}{4}\)-\(\frac{21}{4}\))
=2[(x-\(\frac{3}{2}\))2-\(\frac{21}{4}\)]
2(x-\(\frac{3+\sqrt{21}}{2}\))(x-\(\frac{3-\sqrt{21}}{2}\))
b,x+4\(\sqrt{x}\)+3
=x+3\(\sqrt{x}\)+\(\sqrt{x}\)+3
=\(\sqrt{x}\)(\(\sqrt{x}\)+3) +(\(\sqrt{x}\)+3)
=(\(\sqrt{x}\)+1)(\(\sqrt{x}\)+3)
c,x-5\(\sqrt{x}\)-6
=x-6\(\sqrt{x}\)+\(\sqrt{x}\)-6
=\(\sqrt{x}\)(\(\sqrt{x}\)-6)+(\(\sqrt{x}\)-6)
=(\(\sqrt{x}\)+1)(\(\sqrt{x}\)-6)
d,x+5\(\sqrt{x}\)-14
=x+7\(\sqrt{x}\)-2\(\sqrt{x}\)-14
=\(\sqrt{x}\)(\(\sqrt{x}\)+7)-2(\(\sqrt{x}\)+7)
=(\(\sqrt{x}\)-2)(\(\sqrt{x}\)+7)
Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$
$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$
$=(\sqrt{x}-2)(\sqrt{x}-3)$