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14 tháng 12 2020

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)

 

25 tháng 8 2016

a) 5x2 ( 3x2 -7x+2)-15x(x-3)

=15x4-35x3+10x2-15x2+45x

=15x4-35x3-5x2+45x

c) (x+3)(x-3)(x-2)(x+1)

=(x2-9)(x2+x-2x-2)

=(x2-9)(x2-x-2)

=x4-x3-2x2-9x2+9x+18

=x4-x3-11x2+9x+18

d)(2x+1)2+(4x-1)2+2(2x+1)(4x+1)

=2x2+4x+1-16x2-8x+1

=2x2+4x+1-16x2-8x+1+16x2-4x+8x-2

=2x2+7

e) (2x2-3x)(5x2-2x+1)-10x2(x+3)

=10x4 -4x3+2x2-15x3+6x2-3 -10x2-30x

=10x4-19x3-2x2-30x-3

26 tháng 8 2016

thanks bn nka

4 tháng 8 2017

Ta có : 6x2 - 11x + 3 

= 6x2 - 2x - 9x + 3

= (6x2 - 2x) - (9x - 3)

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

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

4 tháng 8 2017

K MIK NHA BẠN !!!!!!!!!!

bÀI 1 

bÀI 2 : 

Bài 3 :

Bài 4: 

5,

6, 

7, 

8,

9, 

10,

11,

12,

13,

K MIK NHA BẠN !!!!!!!!!!

\(-4x^5\left(x^3-4x^2+7x-3\right)\)

\(=-4x^8+16x^7-28x^6+12x^5\)

b) \(3x^4\left(-2x^3+5x^2-\frac{2}{3}x+\frac{1}{3}\right)\)

\(=-6x^7+15x^6-2x^5+x^4\)

11 tháng 8 2016

đáp án câu a và b có rút gọn được không?

1 tháng 8 2020

\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)

\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)

\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)

\(=-165+22-108=22-273=-251\)

\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)

\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)

\(=3-7.0+5.\left(-3\right)=3-15=-12\)

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

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

\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)

1 tháng 8 2020

\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)

\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)

\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)

\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)

\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)

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

\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)

\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)

\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)

6 tháng 7 2018

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{\left(3x^3-3x^2\right)-\left(4x^2-4x\right)+\left(x-1\right)}{\left(2x^3-2x^2\right)+\left(x^2-x\right)-\left(3x-3\right)}=\frac{\left(x-1\right).\left(3x^2-4x+1\right)}{\left(x-1\right).\left(2x^2+x-3\right)}\\ \)

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

6 tháng 7 2018

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)

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

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

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

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

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

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

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

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

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