1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

\(=\left(x+y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3\)

\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x-y\right)^3\)

\(=x^3-3x^2y+3xy^2-y^3\)

2.

\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3+y^3\)

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

a) $(2x-1)^2-2(2x-3)^2+4$

$=4x^2-4x+1-2(4x^2-12x+9)+4$

$=4x^2-4x+1-8x^2+24x-18+4$

$=-4x^2+20x-13$

b) $(3x+2)^2+2(2+3x)(1-2y)+(2y-1)^2$

Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:

$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$

$=[(3x+2)-(2y-1)]^2$

$=(3x-2y+3)^2$

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

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

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

$=4x^2-4x+1-2(4x^2-12x+9)+4$

$=4x^2-4x+1-8x^2+24x-18+4$

$=-4x^2+20x-13$

Vì $2+3x=3x+2$ và $1-2y=-(2y-1)$:

$=(3x+2)^2-2(3x+2)(2y-1)+(2y-1)^2$

$=[(3x+2)-(2y-1)]^2$

$=(3x-2y+3)^2$

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

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

d) $(x-1)^3+3x(x-1)^2+3x^2(x-1)+x^3$

Đặt $a=x-1,\ b=x$:

$=a^3+3a^2b+3ab^2+b^3$

$=(a+b)^3$

$=[(x-1)+x]^3$

$=(2x-1)^3$

Dùng $(a+b)(a^2-ab+b^2)=a^3+b^3$:

$=(2x)^3+(3y)^3$

$=8x^3+27y^3$

$=x^3-y^3-(x^3+y^3)$

$=-2y^3$

Dùng $(a-b)(a^2+ab+b^2)=a^3-b^3$:

$=(x^2)^3-(2y)^3-x^3(x^3-y^3)+8y^3$

$=x^6-8y^3-x^6+x^3y^3+8y^3$

$=x^3y^3$

Lời giải:
a. Vì $x,y$ tỉ lệ thuận nên đặt $y=kx$. Ta có:

$y_1=kx_1$ hay $\frac{1}{2}=k.2\Rightarrow k=\frac{1}{4}$. Vậy $y=\frac{1}{4}x$

$y_2=kx_2=\frac{1}{4}x_2=\frac{1}{4}.3=\frac{3}{4}$

b.

Vì $x,y$ tỉ lệ nghịch nên đặt $xy=k$.

$x_1y_1=k=x_2y_2$

$\Leftrightarrow \frac{1}{2}.4=x_2.(-4)$

$\Leftrightarrow x_2=\frac{-1}{2}$

a) \(A=\left(x+2\right)\left(x^2-2x+4\right)-x^3+2\)

\(A=x^3+8-x^3+2\)

\(A=10\)

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

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

\(B=x^3-1-x^3-1\)

\(B=-2\)

c) \(C=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(y-3x\right)\left(y^2+3xy+9x^2\right)\)

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

\(C=8x^3-y^3+y^3-27x^3\)

\(C=-19x^3\)

a)

\(A=\left(x+2\right)\left(x-2\right)\left(x-2\right)-x^3+2\\ =\left(x^2-4\right)\left(x-2\right)-x^3+2\\ =x^3-2x^2-4x+8-x^3+2\\ =-2x^2-4x+10\)

b)

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

c)

\(C=\left(2x\right)^3-y^3+\left(y\right)^3-\left(3x\right)^3\\ =8x^3-y^3+y^3-27x^3\\ =-19x^3\)

\(A=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x=x^2+4x\\ B=\left(x^2+xy+y^2\right)\left(x-y\right)=x^3-y^3\)

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

x tỉ lệ thuận với y nên \(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x_1}{y_1}=\frac{x_2}{y_2}=\frac{x_1+x_2}{y_1+y_2}=\frac{8}{2}=4\)

\(\Rightarrow x_1=4y_1\)

Công thức biểu diễn x theo y là x = 4y

a: \(2\left(2x+x^2\right)-x^2\left(x+2\right)+\left(x^3-4x+3\right)\)

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

=3

=>Đúng

b: \(x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

\(=x^3+x_{}^2+x-x^3-x^2-x+5\)

=5

=>Đúng

c: \(3x\left(x-2\right)-5x\left(x-1\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x_{}^2+5x-8x^3+24\)

=-x+24

=>Sai

d: \(2y\left(y^2+y+1\right)-2y^2\left(y+1\right)-2\left(y+10\right)\)

\(=2y^3+2y^2+2y-2y^3-2y^2-2y-20\)

=-20

=>Đúng