Bài 5: Phép cộng các phân thức đại số

a) (2x - 7)/4x + (3 + 4x)/4x

= (2x - 7 + 3 + 4x)/4x

= (6x - 4)/4x

= (3x - 2)/2x

b) (9x - 5)/(x² + x) + (10 - 4x)/(x² + x)

= (9x - 5 + 10 - 4x)/[x(x + 1)]

= (5x + 5)/[x(x + 1)]

= 5(x + 1)/[x(x + 1)]

= 5/x

c) (3x - 9)/(x - 1) : (x² - 9)/(x² - 2x + 1)

= 3(x - 3)/(x - 1) . (x - 1)²/[(x - 3)(x + 3)

= 3(x - 1)/(x + 3)

Lời giải:

Gọi tổng trên là $A$. Ta có:

$A=\frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+\frac{(x+4)-(x+3)}{(x+3)(x+4)}+\frac{(x+5)-(x+4)}{(x+4)(x+5)}$

$=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}$

$=\frac{1}{x+1}-\frac{1}{x+5}=\frac{4}{(x+1)(x+5)}$

c: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{\left(x+3\right)\left(x+2\right)}{\left(x+3\right)\cdot2x}=\dfrac{x+2}{2x}\)

b: \(=\dfrac{x+3-x-4}{x-2}=\dfrac{-1}{x-2}\)

\(\dfrac{5xy-4y}{2x^2y^3}+\dfrac{3xy+4y}{2x^2y^3}=\dfrac{5xy-4y+3xy+4y}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)

\(=\dfrac{x^2-2x+1+9x}{3\left(x+1\right)\left(x-1\right)}=\dfrac{x^2+7x+1}{3\left(x+1\right)\left(x-1\right)}\)

`[x-1]/[3x+3]+[3x]/[x^2-1]`           `ĐK: x \ne +-1`

`=[(x-1)^2+9x]/[3(x+1)(x-1)]`

`=[x^2-2x+1+9x]/[3(x+1)(x-1)]`

`=[x^2+7x+1]/[3(x+1)(x-1)]`

\(\left(x+3\right)^2-x\left(x-4\right)=21\)

\(\Rightarrow x^2+6x+9-x^2-4x=21\)

\(\Rightarrow2x-9=21\)

\(\Rightarrow2x=30\)

\(\Rightarrow x=15\)

Vậy \(x=15\)

\(\left(x+3\right)^2-x\left(x+4\right)=21\)

\(x^2+6x+9-\left(x^2+4x\right)=21\)

\(x^2+6x+9-x^2-4x=21\)

\(\left(x^2-x^2\right)+\left(6x-4x\right)+9=21\)

\(2x+9=21\)

\(2x\)       \(=21-9=12\)

  \(x\)       \(=12:2=6\)

=>x(x^2+6x+9)-x^2-4x=21

=>x^3+6x^2+9x-x^2-4x-21=0

=>x^3+5x^2+5x-21=0

hay \(x\in\left\{1,46\right\}\)

a)\(3(x^2-7)-x(3x+5)=3x^2-21-3x^2-5x=-5x-21\)

b)\(Đk:xy\ne0\)

\(\dfrac{12x^2y^2-6xy^2}{3xy}+2y=\dfrac{3xy\left(4xy-2y\right)}{3xy}+2y=4xy-2y+2y=4xy\)

c)\(Đk:x\ne\pm1\)

\(\dfrac{4}{x+1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x-1\right)+8}{\left(x+1\right)\cdot\left(x-1\right)}=\dfrac{4x+4}{\left(x+1\right)\cdot\left(x-1\right)}=\dfrac{4}{x-1}\)

\(=\dfrac{4x^3-2x^2+6x^2-3x+10x-5}{2x-1}=2x^2+3x+5\)

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