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

b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)

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

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

c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

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

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

\(=ax\left(x-a\right)\)

\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)

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

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

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

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

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

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

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

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

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)

a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)

Thay x = 15 vào bt A ta có

A = 9 . 15 = 135

b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)

Thay x = -1/5 ; y = - 1/2 vào bt B ta có

\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(=9x^2y^2-xy^3-8x^3\)

Thay x = 1/2 ; y = 2 vào bt C ta có

\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)

d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)

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

\(\left|x\right|=2\Rightarrow x=\pm2\)

Thay x = 2 vào bt D có

\(D=12.4+12.2-3=69\)

Thay x = - 2 vào bt D ta có

\(D=12.4-12.2-3=21\)

\(A,xy\left(2x^2-3\right)-x^2\left(5xy+y\right)+x^2y\\ =2x^3y-3xy-5x^3y-x^2y+x^2y\\ =\left(2x^3y-5x^3y\right)+\left(-x^2y+x^2y\right)-3xy\\ =-3x^3y-3xy\)

\(B,3xyz\left(y-2\right)-5yz\left(1-y\right)-8z\left(y^2-3\right)\\ =3xy^2z-6xyz-5yz+5y^2z-8y^2z+24z\\ =3xy^2z-6xyz+\left(5y^2z-8y^2z\right)-5yz+24z\\ =3xy^2z-6xyz-3y^2z-5yz+24z\)

\(A=\dfrac{2x}{x\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3}{x-y}\)

\(=\dfrac{2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)

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

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

a) \(x\left(xy+1\right)+y\left(xy-1\right)-xy\left(x+y\right)\)

\(=X^2y+x+xy^2-y-x^2y-xy^2\)

\(=x-y\)

a, \(x\left(xy+1\right)+y\left(xy-1\right)-xy\left(x+y\right)\)

\(=x^2y+x+xy^2-y-x^2y-xy^2\)

\(=x-y\)

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

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

\(=-2x^2-2\)

a) Ta có: \(3x\left(2x-4\right)-\left(6x-1\right)\left(x+2\right)=25\)

\(\Rightarrow6x^2-12x-\left(6x^2+12x-x-2\right)=25\)

\(\Rightarrow6x^2-12x-6x^2-12x+x+2=25\)

\(\Rightarrow-23x+2=25\)

\(\Rightarrow-23x=25-2-23\)

\(\Rightarrow x=23:\left(-23\right)=-1\)

Vậy x = -1

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

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

\(=\left(x-y\right)2x^2\)

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