a: =>a(x+1)(x+2)+bx(x+2)+cx(x+1)=1

=>a(x^2+3x+2)+bx^2+2bx+cx^2+cx=1

=>ax^2+3ax+2a+bx^2+2bx+cx^2+cx=1

=>x^2(a+b+c)+x(3a+2b+c)+2a=1

=>a+b+c=0 và 3a+2b+c=0 và a=1/2

=>a=1/2; b+c=-1/2; 2b+c=-3/2

=>b=-1; c=1/2; a=1/2

b: =>1=(ax+b)(x-1)+c(x^2+1)

=>x^2*a-a*x+bx-b+cx^2+c=1

=>x^2(a+c)+x(-a+b)-b+c=1

=>a+c=0 và -a+b=0 và -b+c=1

=>a+b=-1 và -a+b=0 và a+c=0

=>a=-1/2; b=-1/2; c=-a=1/2

-ĐKXĐ: \(x\ne\pm1\)

\(\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

\(=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(x^2-x+1\right)}{1+x}-x\right)\right]\)

\(=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(x^2+x+1+x\right)\left(x^2-x+1-x\right)\right]\)

\(=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(x+1\right)^2\left(x-1\right)^2\right]\)

\(=\dfrac{x\left(x-1\right)^2\left(x+1\right)^2}{1+x^2}.\dfrac{1}{\left(x+1\right)^2\left(x-1\right)^2}\)

\(=\dfrac{x}{x^2+1}\)

\(\Rightarrow m=\dfrac{x}{x^2+1}\)

-Khi \(x< 0\), mà \(x^2+1>0\forall x\)

\(\Rightarrow m=\dfrac{x}{x^2+1}< 0\).

\(\Rightarrow m< 0\)

-Vậy khi \(m< 0\) và \(m\ne\dfrac{-1}{2}\) thì \(x< 0\)

a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)

\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)

\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)

c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)

\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)

\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)

\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)

\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)

d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)

\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)

\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)

\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)

\(=\dfrac{x}{x+y}\)

a.

\(\Leftrightarrow x^2+2\left(m-1\right)x+m^2+3m+5\ne0\) ; \(\forall x\)

\(\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m^2+3m+5\right)< 0\)

\(\Leftrightarrow-5m-4< 0\)

\(\Leftrightarrow m>-\dfrac{4}{5}\)

b. 

\(\Leftrightarrow x^2+2\left(m-1\right)x+m^2+m-6\ge0\) ;\(\forall x\)

\(\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m^2+m-6\right)\le0\)

\(\Leftrightarrow-3m+7\le0\)

\(\Rightarrow m\ge\dfrac{7}{3}\)

c.

\(x^2-2\left(m+3\right)x+m+9>0\) ;\(\forall x\)

\(\Leftrightarrow\Delta'=\left(m+3\right)^2-\left(m+9\right)< 0\)

\(\Leftrightarrow m^2+5m< 0\Rightarrow-5< m< 0\)

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