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7 tháng 8 2020

a, (x4-2x3+2x-1):(x2-1) = \(\frac{\left(x^4-1\right)-\left(2x^3-2x\right)}{x^2-1}\) 

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

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

b, (8x3-6x2-5x+3):((4x+3) 

16 tháng 12 2016

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

b ) \(\left(4x^4y^4-12x^2y^2\right):4x^2y^2=x^2y^2-3\)

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

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

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

16 tháng 12 2016

a)   .......=x2-x-2

b)   .........=x2y2-3

c) .......=(3x2-1+x2+1)/2x=4x2/2x=2x

d) x/(x-1)+(-2x)/(x-1)+1/(x-1)=(x2-2x+1)/(x-1)=(x-1)2/(x-1)=x-1

e)...

x-y=4

=> x2-2xy+y2=16

 <=> 106-2xy =16  (vì x2+y2 =106)

=>xy=(106-16)/2=45

ta có   x3 -y=(x-y)(x2+xy+y)

=4(106+45)=604

28 tháng 8 2016

1.a) \(\Leftrightarrow\) 3x+10-2x =0

  \(\Leftrightarrow\text{ 3x-2x=-10}\)

   \(\Leftrightarrow x=-10\)

b) coi lại có thiếu ngoặc ko nhé

cứ nhân vào dấu ngoặc rồi làm như thường

       

24 tháng 12 2017

https://giaibaitapvenha.blogspot.com/2017/12/en-voi-do-homework-for-you-e-trai.html

a) Ta có: \(\left(x^2-1\right)\left(x^2+2x\right)\)

\(=x^4+2x^3-x^2-2x\)

b) Ta có: \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

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

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

\(=18x^2-6x^3+3x-x^2-6+2x\)

\(=-6x^3+17x^2+5x-6\)

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

\(=x^3+3x^2-5x+3x^2+9x-15\)

\(=x^3+6x^2+4x-15\)

d) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)\)

\(=x^3+1\)

e) Ta có: \(\left(2x^3-3x-1\right)\left(5x+2\right)\)

\(=10x^4+4x^3-15x^2-6x-5x-2\)

\(=10x^4+4x^3-15x^2-11x-2\)

f) Ta có: \(\left(x^2-2x+3\right)\left(x-4\right)\)

\(=x^3-4x^2-2x^2+8x+3x-12\)

\(=x^3-6x^2+11x-12\)

g) Ta có: \(\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)

\(=12x^2+4x-3x-1-5x^2+15x-\left(x^2-7x+12\right)\)

\(=7x^2+16x-1-x^2+7x-12\)

\(=6x^2+23x-23\)

h) Ta có: \(\left(5x-2\right)\left(x+1\right)-3x\left(x^2-x-3\right)-2x\left(x-5\right)\left(x-4\right)\)

\(=5x^2+5x-2x-2-3x^3+3x^2+9x-2x\left(x^2-9x+20\right)\)

\(=-3x^3+8x^2+12x-2-2x^3+18x^2-40x\)

\(=-5x^3+26x^2-28x-2\)

a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)

b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)

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

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

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

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

d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)

\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)