Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

a,

$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$

b,

$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$

c,

$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$

d,

$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$

$=3x^2+x$

e,

$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$

f,

$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$

$=\frac{23}{4}xy^2$

a: =3x^3-15x^2+21x

b: =-x^3+6x^2+5x-4x^2-24x-20

=-x^3+2x^2-19x-20

c: =9x^2+15x-3x-5-7x^2-14

=2x^2+12x-19

d: =10x^2-4x+2/3

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,\(2xy.3x.2y^3=\left(2.3.2\right)\left(xx\right)\left(yy^3\right)=12x^2y^4\)

b, \(x\left(x^2-2x+5\right)=x^3-2x^2+5x\)

c, \(\frac{3x^2-6x}{3x}=x-2\)

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

\(x^4.\frac{2x^2-x-1}{3}\)

đúng k bn

đúng rồi bạn Thanh Ngân ơi

Giups mình nhé

\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)