Ôn tập: Phân thức đại số

Bài 5:

a) Ta có: 

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

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

Mà MSC là: \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)

\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow\left(x-a\right)\left(x-b\right)=\left(x-2\right)\left(x-3\right)\)

Vậy \(a=2,b=3\)

b) Hai phân thức đó là:

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

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

đề ạ  

a: 1/x-2=1/x-2

2/2x-4=2/2(x-2)=1/x-2

3/3x-6=3/3(x-2)=1/x-2

b: 1/x+4=(2x-8)/2(x+4)(x-4)

1/2x+8=(x-4)/2(x+4)(x-4)

\(\dfrac{3}{x-4}=\dfrac{3\cdot2\cdot\left(x+4\right)}{2\cdot\left(x-4\right)\left(x+4\right)}=\dfrac{6x+24}{2\left(x-4\right)\left(x+4\right)}\)

c: 1/x^2-1=1/(x-1)(x+1)

2/x-1=2(x+1)/(x-1)(x+1)=(2x+2)/(x-1)(x+1)

2/x+1=2(x-1)/(x+1)(x-1)=(2x-2)/(x-1)(x+1)

d: 1/2x=(x-2)/2x(x-2)

2/x-2=2*2x/2x(x-2)=4x/2x(x-2)

3/2x(x-2)=3/2x(x-2)

a: 1/x^2y=1/x^2y

3/xy=3x/x^2y

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

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

a: \(\dfrac{x^2-4x+4}{x^2-2x}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}=\dfrac{\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)}\)

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

b: \(\dfrac{x^3-2^3}{x^2-4}=\dfrac{x^2+2x+4}{x+2}\)

3/x+2=3/x+2

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

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

\(\dfrac{x^2+4xy+4y^2}{x+2y}=\dfrac{\left(x+2y\right)^2}{x+2y}=x+2y\left(đpcm\right)\)

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

=(x+1)/(2x-1)

Đề ạ

`a, = (3 - 6x)/(2x^3 - x^2 + 2x - 1)`

`= (3(1-2x))/(x^2(2x-1) + 2x-1)`

`= (3(1-2x))/((x^2+1)(2x-1))`

`= -3/(x^2+1)`

`b,` Khi `x = 3` thì GTBT là:

`-3/(3^2+1) = -3/10`

`c,` Do `x^2+1 >= 0+1=1>0 forall x in RR`.

`-3 < 0`

Nên `A` luôn âm

Đề ạ  

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

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

b: Khi x=3 thì \(A=\dfrac{-3}{3^2+1}=-\dfrac{3}{10}\)

c: x^2+1>=0

=>3/x^2+1>=0

=>-3/x^2+1<=0

=>A<=0(ĐPCM)

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)