phân tích thành nhân tử
\(12-\sqrt{x}-x\)
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a: =(căn a-3)^2-b^2
=(căn a-3-b)(căn a-3+b)
b: \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
c: \(x-7\sqrt{x}+12=x-3\sqrt{x}-4\sqrt{x}+12=\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)\)
d: x*căn x-64
=(căn x)^3-4^3
=(căn x-4)(x+4căn x+16)
\(a-6\sqrt{a}+9-b^2\\ =\left(\sqrt{a}+3\right)^2-b^2\\ =\left(\sqrt{a}+3-b\right)\left(\sqrt{a}+3+b\right)\)
\(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
\(x-7\sqrt{x}+12\\ =x-4\sqrt{x}-3\sqrt{x}+12\\ =\sqrt{x}\left(\sqrt{x}-4\right)-3\left(\sqrt{x}-4\right)\\ =\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\)
\(x\sqrt{x}+64\\ =\sqrt{x^3}+4^3\\ =\left(\sqrt{x}\right)^3+4^3\\ =\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)\)
a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)
b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)
c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
\(12-\sqrt{x}-x\)
\(=12+3\sqrt{x}-4\sqrt{x}-x\)
\(=3.\left(4+\sqrt{x}\right)-\sqrt{x}\left(4+\sqrt{x}\right)\)
\(=\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
\(x+7\sqrt{x}+10=\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)\)
a) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
b) \(x-\sqrt{x}-2=\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+1\right)\)
c) \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(12-\sqrt{x}-x\)
ĐK : x ≥ 0
\(=12-4\sqrt{x}+3\sqrt{x}-x\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(3-\sqrt{x}\right)\left(4+\sqrt{x}\right)\)
\(12-\sqrt{x}-x=\left(12-4\sqrt{x}\right)+\left(3\sqrt{x}-x\right)\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)