\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

ngất

a)

(x+y)^2-9x^2 = (x+y-9x)(x+y+9x)=(y-8x)(10x+y)

b,x^2+4-y^2+4x=(x²-y²)+(4+4x)=(x-y)(x+y)+4(1+x)

\(a,3x-6y=3\left(x-2y\right)\)

\(b,\frac{2}{5}x^2+5x^3+x^2y=x^2\left(\frac{2}{5}+5x+y\right)\)

Bài giải:

a) 3x - 6y = 3 . x - 3 . 2y = 3(x - 2y)

b) $\frac{2}{5}$x² + 5x³ + x²y = x² ($\frac{2}{5}$ + 5x + y)

c) 14x²y – 21xy² + 28x²y² = 7xy . 2x - 7xy . 3y + 7xy . 4xy = 7xy(2x - 3y + 4xy)

d) $\frac{2}{5}$x(y - 1) - $\frac{2}{5}$y(y - 1) = $\frac{2}{5}$(y - 1)(x - y)

e) 10x(x - y) - 8y(y - x) =10x(x - y) - 8y[-(x - y)]

= 10x(x - y) + 8y(x - y)

= 2(x - y)(5x + 4y)

a,\(3x-6y=3\left(x-2y\right)\)

b,\(x^2(\dfrac{2}{5}+5x+y)\)

c,\(7xy\left(2x-3y+4xy\right)\)

d,\(\dfrac{2}{5}x\left(y-1\right)-\dfrac{2}{5}y\left(y-1\right)\)

=\(\dfrac{2}{5}\left(y-1\right)\left(x-y\right)\)

e,\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)\)

\(2\left(x-y\right)\left(5x+4y\right)\)

Bài 1:

a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)

b: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

bài 2:

a: Đề thiếu vế phải rồi bạn

b: \(x^3-13x=0\)

=>\(x\left(x^2-13\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)

Bài 1:

a, $3x-6y$

$=3(x-2y)$

b, $14x^2y-21xy^2+28x^2y^2$

$=7xy(2x-3y+4xy)$

c, $10x(x-y)-8y(y-x)$

$=10x(x-y)-8y[-(x-y)]$

$=10x(x-y)+8y(x-y)$

$=(x-y)(10x+8y)$

$=2(x-y)(5x+4y)$

Bài 2:

a, Đề thiếu rồi bạn nhé.

b, \(x^3-13x=0\)

\(\Rightarrow x\left(x^2-13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)

Bài 1:

a) \(3x-6y=3\left(x-2y\right)\)

b)  \(\frac{2}{5}x^2+x^3+x^2y\)

\(=x^2\left(\frac{2}{5}+x+y\right)\)

c) \(14x^2y-21xy^2+28x^2y^2\)

\(=xy\left(14x-21y+28xy\right)\)

d) \(\frac{2}{3}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)

\(=\left(y-1\right)\left(\frac{2}{3}x-\frac{2}{5}y\right)\)

e) \(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

f) \(4x^2-25\)

\(=\left(2x\right)^2-5^2\)

\(=\left(2x-5\right)\left(2x+5\right)\)

g) \(6x-9-x^2\)

\(=-x^2+6x-9\)

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

h) \(x^3+y^3+z^3-3xyz\)

Đề sai ?? ko tách đc nữa

Bài 2:

a) \(15.91,5+150.0,85\)

\(=15.\left(91,5+8,5\right)\)

\(=15.100\)

\(=1500\)

b) \(x\left(x-1\right)-y\left(1-x\right)\)

\(=x\left(x-1\right)+y\left(x-1\right)\)

\(=\left(x-1\right)\left(x+y\right)\)

\(=2000.4000\)

\(=8000000\)

c) \(87^2+73^2-27^2-13^2\)

\(=\left(87^2-27^2\right)+\left(73^2-13^2\right)\)

\(=\left(87-27\right)\left(87+27\right)+\left(73-13\right)\left(73+13\right)\)

\(=60.114+60.86\)

\(=60.\left(144+86\right)\)

\(=60.230\)

\(=82800\)

d) \(73^2-27^2\)

\(=\left(73-27\right)\left(73+27\right)\)

\(=46.100\)

\(=4600\)

e) f) tương tự

Bài 3:

a) \(5x\left(x-2000\right)-x+2000=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)

Vậy ...

b) \(x^3-13x=0\)

\(\Leftrightarrow x\left(x^2-13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}}\)

Vậy ...

c) \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy

d) \(2-25x^2=0\)

\(\Leftrightarrow\left(\sqrt{2}-5x\right)\left(\sqrt{2}+5x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2}-5x=0\\\sqrt{2}+5x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\sqrt{2}\\5x=-\sqrt{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{cases}}\)

Vậy ...

h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)

k) \(=\left(x+2\right)\left(3x-5\right)\)

l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)

m) \(=7xy\left(2x-3y+4xy\right)\)

n) \(=2\left(x-y\right)\left(5x-4y\right)\)