\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

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

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

=x+y-z

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

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

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

1, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\)

\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)

2, \(\dfrac{x^2+4x+3}{2x+6}\)

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

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

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

3, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

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

5, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

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

\(A=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-b\right)}=\dfrac{2a\left(x-1\right)^2}{5b\left(1-b\right)}\)

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

B=\(\frac{5\left(x-y\right)-3\left(x-y\right)}{10\left(x-y\right)}\)

B=\(\frac{\left(x-y\right)\left(5-3\right)}{10\left(x-y\right)}\)

B= \(\frac{\left(x-y\right)2}{10\left(x-y\right)}\)

B= 5

vậy B=5

\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)

\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)

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

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

\(\frac{2ax^2-4ax+2a}{5b-5bx^2}\)

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

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

tôi cũng cung thiên yết nè nhưng lại là cậu bé mà thiên yết hợp với cung gì nhất vậy add friend nha

Ta có: \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

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

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

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

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

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}=\dfrac{x^2-xy+y^2}{x^3+xy^2-x^2y-y^3}\)