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4 tháng 12 2018

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

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

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

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

\(=\frac{x^2-4}{x^2+5x+2x+10}\)

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

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

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

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

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

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

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

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

26 tháng 5 2023

\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)

\(=0-11x+24\)

\(=-11x+24\)

\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)

\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)

\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)

\(=0+0+5\)

\(=5\)

23 tháng 12 2021

\(a,\dfrac{x^3-x}{3x+3}=\dfrac{x\left(x^2-1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)\left(x+1\right)}{3\left(x+1\right)}=\dfrac{x\left(x-1\right)}{3}\\ b,\dfrac{x^2+3xy}{x^2-9y^2}=\dfrac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\dfrac{x}{x-3y}\\ c,\dfrac{x^2+4x+4}{3x+6}=\dfrac{\left(x+2\right)^2}{3\left(x+2\right)}=\dfrac{x+2}{3}\)

24 tháng 7 2020

a, Điều kiện xác định: x<>0

b, Điều kiện xác định: x <> -1/3

c, Điều kiện xác định: x<>2

d, Điều kiện xác định: a<>0 và b<>0; b<>2a

A : không rút gọn được

\(B=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{3x\left(4x^2+3\right)+4x^2+3}=\frac{\left(4x^2+3\right)\left(x-2\right)}{\left(4x^2+3\right)\left(3x+1\right)}=\frac{x-2}{3x+1}\)

\(C=\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)

\(D=\frac{a^3+b^3}{a^3+\left(a-b\right)^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+a-b\right)\left(a^2-a^2+ab+a^2-2ab+b^2\right)}\)\(=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(2a-b\right)\left(a^2-ab+b^2\right)}=\frac{a+b}{2a-b}\)

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

22 tháng 10 2023

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

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

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

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

30 tháng 11 2016

a. 2x

b.\({3x}\over x^2-1\)