Bài 3: Rút gọn phân thức

\(\dfrac{10x^2y^3}{15xy}=\dfrac{2xy^2}{3}\)

\(\dfrac{10x^2y^3}{15xy}=\dfrac{10x^2y^3:5xy}{15xy:5xy}=\dfrac{2xy^2}{3xy}\)

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

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

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

29: =(x-2)(x-3)/(x-2)(x+2)=(x-3)/(x+2)

27: =(x-7)(x+6)/x(x-7)=(x+6)/x

Coi như tất cả các câu đều trên điều kiện xác định.

`21)[x^3-x]/[x^3+2x^2+x]`

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

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

`=[x-1]/[x+1]`

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`22)[x^2-2x+1]/[x^2-x]`

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

`=[x-1]/x`

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`23)[x^2-9y^2]/[x^2-6xy+9y^2]`

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

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`24)[3x^3-27x]/[-6x^2+18x]`

`=[3x(x^2-9)]/[-6(x-3)]`

`=[3x(x-3)(x+3)]/[-6(x-3)]=[-x(x+3)]/2=[-x^2-3x]/[2`

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`25)[x^2-4x+4]/[2x^2-4x]`

`=[(x-2)^2]/[2x(x-2)]=[x-2]/[2x]`

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`26)[x^2-2x]/[x^2-4]`

`=[x(x-2)]/[(x-2)(x+2)]`

`=x/[x+2]`

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`27)[x^2-x-42]/[x^2-7x]`

`=[(x-7)(x+6)]/[x(x-7)]`

`=[x+6]/x`

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`28)[3x^2-x-2]/[x^2-1]`

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

`=[3x+2]/[x+1]`

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`29)[x^2-5x+6]/[x^2-4]`

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

`=[x-3]/[x+2]`

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`30)[x^2-3x+2]/[x^2-x]`

`=[(x-2)(x-1)]/[x(x-1)]`

`=[x-2]/x`

16: \(=\dfrac{5\left(x^2+2xy+y^2\right)}{3\left(x-y\right)\left(x+y\right)}=\dfrac{5\left(x+y\right)^2}{3\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{5x+5y}{3x-3y}\)

17: \(=\dfrac{x\left(x^2-6x+9\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-3x}{x+3}\)

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

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

20: \(=\dfrac{\left(2x-1\right)^2}{3y\cdot\left(2x-1\right)}=\dfrac{2x-1}{3y}\)

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

Em nhập lại câu hỏi em hi

bn tách từng câu ra ạ

a.

ĐKXĐ: \(x\ne\dfrac{1}{2}\)

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

b.

\(x=3\Rightarrow A=\dfrac{-3}{3^2+1}=-\dfrac{3}{10}\)

c.

Ta có: \(x^2+1>0;\forall x\)

\(\Rightarrow\dfrac{-3}{x^2+1}< 0;\forall x\)

Vậy \(A< 0;\forall x\ne\dfrac{1}{2}\)

10.

\(x^3+8+\left(x+2\right)\left(x+4\right)=\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)\left(x+4\right)\)

\(=\left(x+2\right)\left(x^2-x+8\right)\)

11.

\(2x\left(3x-5\right)-10+6x=2x\left(3x-5\right)+2\left(3x-5\right)=2\left(x+1\right)\left(3x-5\right)\)

12.

\(\left(5x-4\right)^2-\left(7x\right)^2=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)=-8\left(x+2\right)\left(3x-1\right)\)

13.

\(4x^2+4x+1-9y^2=\left(2x+1\right)^2-\left(3y\right)^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\)

14.

\(x^2-2xy+y^2+4x-4y=\left(x-y\right)^2+4\left(x-y\right)=\left(x-y\right)\left(x-y+4\right)\)

10)

 \(x^3+8-\left(x+2\right)\left(x+4\right)\\ =\left(x+2\right)\left(x^2-2x+4\right)-\left(x+2\right)\left(x+4\right)\\ =\left(x+2\right)\left[\left(x^2-2x+4\right)-\left(x+4\right)\right]\\ =\left(x+2\right)\left(x^2-2x+4-x-4\right)\\ =\left(x+2\right)\left(x^2-3x\right)\\ =x\left(x+2\right)\left(x-3\right)\)

11)

\(2x\left(3x-5\right)-10+6x\\ =6x^2-10x-10+6x\\ =6x^2-4x+10\\ =2\left(3x^2-2x+5\right)\)

12)

\(\left(5x-4\right)^2-49x^2\\ =\left[\left(5x-4\right)-7x\right]\left[\left(5x-4\right)+7x\right]\\ =\left(-4-2x\right)\left(13x-4\right)\\ =\left[-2\left(2+x\right)\right]\left(13x-4\right)\)

14)

\(x^2+4x-2xy-4y+y^2\\ =\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\\ =4\left(x-y\right)+\left(x-y\right)^2\\ =\left(x-y\right)+\left[4\left(x-y\right)\right]\)

13)

\(4x^2+4x-9y^2+1\\ =4x\left(x+1\right)-9y^2+1\)