a) Ta có: \(\frac{5}{{2 - x}} = \frac{{ - 5}}{{x - 2}}\)

\({x^2} - 4{\rm{x}} + 4 = {\left( {x - 2} \right)^2}\)

\(MTC = \left( {x + 2} \right){\left( {x - 2} \right)^2}\)

Nhân tử phụ của x+2 là \({\left( {x - 2} \right)^2}\)

Nhân tử phụ của\({x^2} - 4{\rm{x}} + 4\)  là \({\left( {x - 2} \right)^2}\)

Nhân tử phụ của x - 2 là (x+2)(x−2)

Nhân cả tử và mẫu của mỗi phân thức với nhân tử phụ tương ứng, ta có:

\(\begin{array}{l}\frac{1}{{x + 2}} = \frac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}}\\\frac{{x + 1}}{{{x^2} - 4{\rm{x  -  4}}}} = \frac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}}\\\frac{5}{{2 - x}} = \frac{{ - 5\left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x + 2} \right){{\left( {x - 2} \right)}^2}}}\end{array}\)

b) Ta có: 3x+3y=3(x+y)

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

            \({x^2} + 2{\rm{x}}y + {y^2} = {\left( {x - y} \right)^2}\)

\(MTC = 3\left( {x + y} \right){\left( {x - y} \right)^2}\)

Nhân tử phụ của 3x+3y là: \({\left( {x - y} \right)^2}\)

Nhân tử phụ của \({x^2} - {y^2}\) là: 3(x−y)

Nhân tử phụ của \({x^2} + 2{\rm{x}}y + {y^2}\) là: 3(x+y)

Nhân cả tử và mẫu của mỗi phân thức với nhân tử phụ tương ứng, ta có: 

\(\begin{array}{l}\frac{1}{{3{\rm{x}} + 3y}} = \frac{{{{\left( {x - y} \right)}^2}}}{{3\left( {x + y} \right){{\left( {x - y} \right)}^2}}}\\\frac{{2{\rm{x}}}}{{{x^2} - {y^2}}} = \frac{{6{\rm{x}}\left( {x - y} \right)}}{{3\left( {x + y} \right){{\left( {x - y} \right)}^2}}}\\\frac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}} = \frac{{3\left( {{x^2} - xy + {y^2}} \right)\left( {x + y} \right)}}{{3\left( {x + y} \right){{\left( {x - y} \right)}^2}}}\end{array}\)

Bài 1 . Chia :( x³ + 5x² - 4x - 20) cho ( x² + 3x - 10) ta được x+ 2

Chia :( x³ + 5x² - 4x - 20) cho ( x² + 7x + 10) ta được x - 2

Do đó , ta có :

\(\dfrac{1}{x^2+3x-10}=\dfrac{x+2}{\left(x^2+3x-10\right)\left(x+2\right)}=\dfrac{x+2}{x^3+5x^2-4x-20}\)

Và : \(\dfrac{x}{x^2+7x+10}=\dfrac{x\left(x-2\right)}{\left(x^2+7x+10\right)\left(x-2\right)}=\dfrac{x^2-2x}{x^3+5x^2-4x-20}\)

Bài 2 . a) Ta có :

\(\dfrac{x-1}{x^3+1}\)( giữ nguyên)

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

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

b) Ta có MTC = x²( y - z)²

Ta có :

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

\(\dfrac{y}{x^2\left(y-z\right)^2}\)( giữ nguyên )

\(\dfrac{z}{x^2}=\dfrac{z\left(y-z\right)^2}{x^2\left(y-z\right)^2}\)

MTC=x^2(y-z)^2

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

\(\dfrac{y}{x^2\left(y-z\right)^2}=\dfrac{y}{x^2\left(y-z\right)^2}\)

\(\dfrac{z}{x^2}=\dfrac{z\left(y-z\right)^2}{x^2\left(y-z\right)^2}\)

a: \(\frac{x+y}{x^2\left(y+z\right)}=\frac{\left(x+y\right)\cdot y^2z^2\left(x+z\right)\left(x+y\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{\left(x+z\right)\cdot y^2z^2\left(x+y\right)^2}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

\(\frac{y+z}{y^2\left(x+z\right)}=\frac{\left(y+z\right)\cdot x^2z^2\left(x+y\right)\left(y+z\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{x^2z^2\left(y+z\right)^2\cdot\left(x+y\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

\(\frac{z+x}{z^2\left(x+y\right)}=\frac{\left(z+x\right)\cdot x^2y^2\cdot\left(x+z\right)\left(y+z\right)}{x^2y^2z^2\left(x+y\right)\left(x+z\right)\left(y+z\right)}=\frac{x^2y^2\left(x+z\right)^2\cdot\left(y+z\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

b: \(\frac{5x}{x^2+5x+6}=\frac{5x}{\left(x+2\right)\left(x+3\right)}=\frac{5x\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(\frac{2x+3}{x^2+7x+10}=\frac{2x+3}{\left(x+2\right)\left(x+5\right)}=\frac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(-5=\frac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

1 a

2c

3b

4d

5c

6c