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\(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
\(x^2-6=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)
\(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}.x+\sqrt{3}^2=\left(x+\sqrt{3}\right)^2\)
\(x^2-2\sqrt{5}x+5=x^2-2\sqrt{5}x+\sqrt{5}^2=\left(x-\sqrt{5}\right)^2\)
\(a,2x-2\sqrt{x}=2\sqrt{x}\left(\sqrt{x}-1\right)\\ b,x-\sqrt{x}-6=x-3\sqrt{x}+2\sqrt{x}-6\\ =\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\\ c,4x-4\sqrt{x}+1=\left(2\sqrt{x}\right)^2-2.2\sqrt{x}.1+1^2=\left(2\sqrt{x}+1\right)^2\)
a) \(2x-2\sqrt{x}=2\sqrt{x}\left(\sqrt{x}-1\right)\)
b) \(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
c) \(4x-4\sqrt{x}+1=\left(2\sqrt{x}-1\right)^2\)
a) \(a-5\sqrt{a}\)
\(=\sqrt{a}\left(\sqrt{a}-\sqrt{5}\right)\)
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) \(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12\)
\(=\sqrt{x}\left(\sqrt{y}-4\right)+3\left(\sqrt{y}-4\right)\)
\(=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)
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: \(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12\)
=căn x(căn y-4)+3(căn y-4)
=(căn y-4)(căn x+3)
a3 + b3 + c3 - 3abc = (a + b)3 + c3 - 3abc - 3ab(a + b)
= (a + b + c)(a2 + b2 + 2ab - ac - bc + c2) - 3ab(a + b + c)
= (a + b + c)(a2 + b2 + c2 - ab - ac - bc)
\(a,=\sqrt{xy}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{xy}+1\right)\left(\sqrt{x}-1\right)\\ b,=\sqrt{xy}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{xy}+1\right)\)
\(\Delta=9-4\left(-m^2+m+2\right)=4m^2-4m+1=\left(2m-1\right)^2\)
Pt có 2 nghiệm pb khi \(m\ne\dfrac{1}{2}\)
Do vai trò của 2 nghiệm là như nhau, giả sử: \(\left\{{}\begin{matrix}x_1=\dfrac{3-\left(2m-1\right)}{2}=2-m\\x_2=\dfrac{3+2m-1}{2}=m+1\end{matrix}\right.\)
\(x_1^2+x_2^2=5\Leftrightarrow\left(2-m\right)^2+\left(m+1\right)^2=5\)
\(\Leftrightarrow m^2-m=0\Rightarrow\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\)
