\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

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

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

\(=\dfrac{x-y+z}{x-y-z}\)

x² +y² +z² -2xy-2zx-2yz=(x-y-z)² -4yz=(x-y-z)² - \(2.\sqrt{yz^2}\)=\(\left(x-y-z-2\sqrt{yz}\right)+\left(x-y-z+2\sqrt{yz}\right)\)

x² -2xy - y² -z² =(x-y)² -z² =(x-y-z)(x-y+z)

\(\frac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

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

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

\(=\frac{x-y+z}{x-y-z}\)

\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)

\(=-5x-27\)

\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

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

\(=8x^3+y^3-8x^3+y^3=2y^3\)

\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)

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

\(=z^2\)

a)

=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\) 

=-5x-19

b)

=\(8x^3+y^3-8x^3+y^3\) 

=\(2y^3\) 

c)

=(x+y+z-x-y)\(^2\) +x+y

=\(z^2+x+y\) 

hc tốt

Bài 1:

a: \(\dfrac{xy}{x^2+y^2}=\dfrac{5}{8}\)

=>\(\dfrac{xy}{5}=\dfrac{x^2+y^2}{8}=k\)

=>\(xy=5k;x^2+y^2=8k\)

\(A=\dfrac{8k-2\cdot5k}{8k+2\cdot5k}=\dfrac{-2}{18}=\dfrac{-1}{9}\)

b: Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\)

=>x=a*k; y=b*k; z=c*k

\(B=\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2}\)

\(=\dfrac{k^2\cdot\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

b, (a+b) ^3 - ( a - b)^3- 2 b^3

= ( a +b -a +b) [ ( a+ b)^2 + (a+b)(a-b) + (a-b)^2]  - 2b^3

= 2b( a^2 + 2ab+ b^2 + a^2 - b^2 + a^2 - 2ab+ b^2 ) - 2b^3

= 2b (  3 a^2 + b^2) - 2b^3

= 2b ( 3a^2 + b^2 - b^2)

= 2b.3a^2

=6a^2b

Có lẽ không thể rút gọn thêm được đâu