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Lời giải:
a.
\(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)
\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)
\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)
b.
\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)
\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)
c.
\(\frac{4x^2-3x+5}{x^3-1}\)
\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)
\(-2=\frac{-2(x^3-1)}{x^3-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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xy + 2x + y + 2 = y(x + 1) + 2(x + 1) = (x + 1).(y + 2)
x(x - 1) + x(x + 3) = x(x - 1 + x + 3) = x. ( 2x + 2) = 2x.(x + 1)
\(-4x^2+8x-4=-4\left(x^2-2x+1\right)=-4\left(x-1\right)^2\)
c: \(-4x^2+8x-4\)
\(=-4\left(x^2-2x+1\right)\)
\(=-4\left(x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^4+x^3-8x-8\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x^3+8\right)\left(x+1\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x+1\right)\)
a) \(=x^3\left(x+1\right)-8\left(x+1\right)=\left(x+1\right)\left(x^3-8\right)=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(=y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(y-3\right)\)
c) \(=3\left(x-y\right)-a\left(x-y\right)=\left(x-y\right)\left(3-a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(a+b)3+(c-a)3-(b+c)3=(b+c)((a+b)2-(a+b)(c-a)+(a-c)2)-(b+c)3=(b+c)(a2+b2+2ab+a2-ac-bc+ab+a2-2ac+c2-b2-c2-2bc)=(b+c)(3a2+3ab-3ac-3bc)=3(b+c)(a-c)(a+b)
Yêu cầu đề là j
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