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\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)
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\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)
a) 2x2+6x=2x(x+3)
b) x4+3x3+x+3=(x4+x)+(3x3+3)=x(x3+1)+3(x3+1)=(x+3)(x3+1)
c) 64-x2-y2+2xy=-(x2-2xy+y2)+82=8-(x+y)2=(8+x+y)(8-x-y)
A= (x+5)(x+1)+(x-2)(x2+2xx+4)-(x2+x-2)
A= x2+6x+5+x3-8-x2-x+2
A= x3+(x2-x2)+(6x-x)+(5-8+2)
A= x3+5x-1
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Ta có: \(6x^2-7x+2=6x^2-3x-4x+2\)
\(=\left(6x^2-3x\right)-\left(4x-2\right)\)
\(=3x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(3x-2\right)\)
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\(64-x^2-y^2+2xy=64-\left(x^2-2xy+y^2\right)\)
\(=8^2-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\)
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a: \(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
hay \(x\in\left\{0;\sqrt{3};-\sqrt{3}\right\}\)
b: \(=\dfrac{x^3-3x^2+6x-8}{x-2}=\dfrac{x^2-2x-x^2+2x+4x-8}{x-2}=x^2-x+4\)
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a)lát đề
b)x3+x+2
=x3-x2+2x+x2-x+2
=x(x2-x+2)+(x2-x+2)
=(x+1)(x2-x+2)
c)x4+64
=(x2)2+82+2x2*8-2x2*8
=(x2+8)2-(4x)2
=(x2-4x+8)(x2+4x+8) .
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\(=x^3+4x^2-10x^2-40x+16x+64\\ =\left(x+4\right)\left(x^2-10x+16\right)\\ =\left(x+4\right)\left(x^2-2x-8x+16\right)\\ =\left(x+4\right)\left(x-2\right)\left(x-8\right)\)
Làm ơn nhanh giùm mik nha