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9 tháng 11 2017

a) \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

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

=\(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)

= \(\dfrac{\left(x^2-4\right)\left(x+1\right)}{\left(x^2+7x+10\right)\left(x+1\right)}\)

= \(\dfrac{\left(x+2\right)\left(x-2\right)\left(x+1\right)}{\left[\left(x^2+2x\right)+\left(5x+10\right)\right]\left(x+1\right)}\)

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

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

= \(\dfrac{x-2}{x+5}\)

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

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

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

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

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

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

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

Nhớ tik nha...leuleuleuleuleuleu

10 tháng 11 2017

Rút gọn phân thức

24 tháng 11 2018

\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)

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

Còn bài b bạn tự làm nhé

24 tháng 11 2018

Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)

\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)

\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)

\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)

Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)

\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)

NV
24 tháng 11 2018

a/ \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\dfrac{2x^3-12x^2+18x+5x^2-30x+45}{3x^3-18x^2+27x-x^2+6x-9}\)

\(=\dfrac{2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)}{3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)}=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)

\(=\dfrac{2x+5}{3x-1}\)

b/ \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\dfrac{x^3+3x^2+2x-2x^2-6x-4}{x^3+3x^2+2x+5x^2+15x+10}\)

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

\(=\dfrac{x-2}{x+5}\)

AH
Akai Haruma
Giáo viên
24 tháng 11 2018

Lời giải:

ĐKXĐ:.........

a) \(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-(x^2-3x)-(15x-45)}{3x^3-9x^2-(10x^2-30x)+(3x-9)}\)

\(=\frac{2x^2(x-3)-x(x-3)-15(x-3)}{3x^2(x-3)-10x(x-3)+3(x-3)}=\frac{(x-3)(2x^2-x-15)}{(x-3)(3x^2-10x+3)}\)

\(=\frac{(x-3)[2x(x-3)+5(x-3)]}{(x-3)[3x(x-3)-(x-3)]}=\frac{(x-3)(x-3)(2x+5)}{(x-3)(x-3)(3x-1)}=\frac{2x+5}{3x-1}\)

b)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2(x+1)-4(x+1)}{x^3+x^2+7x^2+7x+10x+10}\)

\(=\frac{(x+1)(x^2-4)}{x^2(x+1)+7x(x+1)+10(x+1)}=\frac{(x+1)(x-2)(x+2)}{(x+1)(x^2+7x+10)}\)

\(=\frac{(x-2)(x+2)}{x^2+7x+10}=\frac{(x-2)(x+2)}{x(x+2)+5(x+2)}=\frac{(x-2)(x+2)}{(x+2)(x+5)}=\frac{x-2}{x+5}\)

23 tháng 11 2017

a)

\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

\(=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+2x^2+6x^2+12x+5x+10}\)

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

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+6x+5\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{x-2}{x+5}\)

b)

\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)

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

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

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

a: \(=\dfrac{x^4+15x+7}{x^4+15x+7}\cdot\dfrac{x}{14x^2+1}\cdot\dfrac{4x^3+4}{2x^3+2}=\dfrac{2x}{14x^2+1}\)

b: \(=\dfrac{x^7+3x^2+2}{x^7+3x^2+2}\cdot\dfrac{x^2+x+1}{x^3-1}\cdot\dfrac{3x}{x+1}\)

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

a: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)

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

c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)

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

d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)

\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)

28 tháng 6 2017

Phép nhân các phân thức đại số

28 tháng 6 2017

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Rút gọn phân thức

3 tháng 7 2018

đề bài kêu làm gì

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