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

câu 1 là (x-y).(6x^2-4y^2+\(\dfrac{1}{2}\)xy) nha 

1. \(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)\)

\(=6x^3-4xy^2+\dfrac{1}{2}x^2y-6x^2y+4y^3-\dfrac{1}{2}xy^2\)

\(=6x^3+4y^3-\dfrac{11}{2}x^2y-\dfrac{9}{2}xy^2\)

===========

2. \(\left(6x-1\right)\left(3+x\right)+\left(2x+5\right)\left(-3x\right)\)

\(=18x+6x^2-3-x-6x^2-15x\)

\(=2x-3\)

Chúc bạn học tốt!

`1)(x-y)(6x^2-4y^2+1/2xy)`

`=6x^3-4xy^2+1/2x^2y-6x^2y+4y^3-1/2xy^2`

`=6x^3-11/2x^2y-9/2xy^2+4y^3`

`2)(6x-1)(3+x)+(2x+5)(-3x)`

`=18+6x^2-x-3-6x^2-15x`

`=-15x+15`

1)\(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)=6x^3+4y^3-\dfrac{9}{2}xy^2-\dfrac{11}{2}x^2y\)

2)\(\left(6x-1\right)\left(3+x\right)+\left(2x+5\right)\left(-3x\right)=\left(6x^2+17x-3\right)+\left(-6x^2-15x\right)=2x-3\)

tick mik nha

2: Ta có: \(\left(6x-1\right)\left(x+3\right)+\left(-3x\right)\left(2x+5\right)\)

\(=6x^2+18x-x-3-6x^2-15x\)

\(=2x-3\)

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: \(=\dfrac{4x-2+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2x-1}{2x\left(2x-1\right)}\)

a, Biến đổi vế trái :

\(VT=x\left(x+1\right)\left(x+2\right)=\left(x^2+x\right)\left(x+2\right)=x^3+3x^2+2x\) 2x

b,\(\left(3x-2\right)\left(4x-5\right)-\left(2x-1\right)\left(6x+2\right)=0\)

\(\Leftrightarrow12x^2-15x-8x+10-\left(12x^2+4x-6x-2\right)=0\)

\(\Leftrightarrow12x^2-23x+10-12x^2+2x+2=0\)

\(\Leftrightarrow12-21x=0\)

\(\Leftrightarrow-21x=-12\)

\(\Leftrightarrow21x=12\)

\(\Leftrightarrow x=\frac{4}{7}\)

c,

a, bạn thêm

Vậy VT=VT(đpcm)

nhé

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

^^

\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)

\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)

c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

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