a,(x^2+2xy+y^2):(x+y) 

= (x+y)² : (x+y)

= x+y

 b,(125x^3+1):(5x+1)

=( 5x +1) (25x² - 5x +1) : (5x+1)

= 25x² -5x +1

    c,(x^2-2xy+y^2):(x-y)

= (x-y)² : (x-y)

=x-y

4x²-(x-2)²=0

<=> (2x)² - (x-2)² = 0

<=> (2x -x+2)(2x+x-2) = 0

<=> \(\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(\dfrac{7}{9}>\dfrac{7}{8}\left(S\right)\)

\(\dfrac{111}{112}=\dfrac{112}{113}\left(S\right)\)

\(\dfrac{3}{5}< \dfrac{5}{7}\left(Đ\right)\)

\(\dfrac{13}{9}=\dfrac{117}{81}\left(Đ\right)\)

\(\dfrac{8}{26}=\dfrac{8:2}{26:2}=\dfrac{4}{13}\)

\(\dfrac{16}{40}=\dfrac{16:8}{40:8}=\dfrac{2}{5}\)

\(\dfrac{19}{57}=\dfrac{19:19}{57:19}=\dfrac{1}{3}\)

\(\dfrac{45}{105}=\dfrac{3}{7}\)

7/4 - (x + 5/3) = -12/5 

<=> x+5/3 = 7/4 + 12/5

<=> \(x+\dfrac{5}{3}=\dfrac{83}{20}\)

<=> x = \(\dfrac{83}{20}-\dfrac{5}{3}=\dfrac{149}{60}\)

\(\dfrac{9}{24}=\dfrac{9:3}{24:3}=\dfrac{3}{8}\)

\(\dfrac{13}{39}=\dfrac{13:13}{39:13}=\dfrac{1}{3}\)

\(\dfrac{35}{49}=\dfrac{35:7}{49:7}=\dfrac{5}{7}\)

\(a.3-\left(\dfrac{-6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\dfrac{1}{4}:2\)

\(=2+\dfrac{1}{8}\)

=\(\dfrac{16}{8}+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

a) \(2xy^2-10xy\)

\(=2xy\left(y-5\right)\)

b) \(x^2+xy-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

c) \(25-x^2-y^2+2xy\)

= \(25-\left(x^2+y^2-2xy\right)\)

= \(25-\left(x-y\right)^2\)

= \(\left(5-x+y\right)\left(5+x-y\right)\)

Để đa thức A chia hết cho đa thứ B thì :

\(-3b+15=0\)

\(< =>-3b=-15\)

\(< =>b=-15:3=5\)

Vậy để đa thức A chia hết cho đa thức B thì b = 5

* Làm theo sơ đồ hoocne

\(A=-3x\left(x-5\right)+5\left(x-1\right)+3x^2\)

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

\(A=20x-5\)

\(B=x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

\(B=2x^3-3x-5x^3-x^2+x^2\)

\(B=-3x^3-3x\)

\(C=5\left(x^2-3x+1\right)+x\left(5x+15\right)+5\)

\(C=5x^2-15x+5+5x^2+15x+5\)

\(C=10x^2+10\)

\(D=4x\left(x^2-x+1\right)-x\left(4x^2-2x-5\right)\)

\(D=4x^3-4x^2+4x-4x^3+2x^2+5x\)

\(D=-2x^2+9x\)