\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)

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

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

\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)

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

\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)

a) x^2+2xy+y^2-16

=(x+y)²-16

=(x+y-4)(x+y+4)

b) 3x^2+5x-3xy-5y

=(3x²-3xy)+(5x-5y)

=3x(x-y)+5(x-y)

=(x-y)(3x+5)

c) 4x^2-6x^3y-2x^2+8x

ko bik hoặc sai đề

d) x^2-4-2xy+y^2

=(x-y)²-4

=(x-y+2)(x-y-2)

e) x^3-4x^2-12x+27

=sai đề

g) 3x^2-18x+27

=3(x²-6x+9)

=3(x-3)²

h) x^2-y^2-z^2-2yz

=x²-(y²+z²+2yx)

=x²-(y+z)²

=(x-y-z)(x+y+z)

k) 4x^2(x-6)+9y^2(6-x)

=4x²(x-6)-9y²(x-6)

=(x-6)(4x²-9y²)

=(x-6)(2x-3y)(2x+3y)

l)6xy+5x-5y-3x^2-3y^2

=(5x-5y)+(-3x²+6xy-3y²)

=5(x-y)-3(x²-2xy+y²)

=5(x-y)-3(x-y)²

=(x-y)(5-3(x-y))

=(x-y)(5-3x+3y)

ĐKXĐ bạn tự tìm nha : )

k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)

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

j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)

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

i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)

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

h, = k,

f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)

a. \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right).2\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\frac{-5}{2}\)

b. \(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{12x.15y^4}{5y^3.8x^3}=\frac{3.3y}{2x^2}=\frac{9y}{2x^2}\)

c. \(\frac{4y^2}{11x^4}.\left(\frac{-3x^2}{8y}\right)=\frac{4y^2.\left(-3x^2\right)}{11x^4.8y}=\frac{-3y}{22x^2}\)

d. \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}=\frac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\frac{x+2}{6}\)

f. \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x+6\right)\left(x-6\right).3}{\left(2x+10\right)\left(6-x\right)}=\frac{-3x-18}{2x+10}\)

g. \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6}=\frac{\left(x^2-9y^2\right).3xy}{x^2y^2.\left(2x-6\right)}=\frac{3x^2-27y^2}{2x^2y-6xy}\)

h. \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right).3x}{x\left(x+4\right).2\left(1-2x\right)}=\frac{3+6x}{2x+8}\)

i. \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right).2\left(a-b\right)\left(a+b\right)}{\left(b-a\right)\left(a+b\right)}=-2a^2-2ab\)

j. \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{\left(x+y\right).3\left(x-y\right)\left(x+y\right)}{\left(y-x\right).x\left(x+y\right)}=\frac{-3x-3y}{x}\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)