\(\begin{cases} x^3 + 4y = y^3 + 16x & (1) \\ 1 + y^2 = 5(1 + x^2) & (2) \end{cases}\)

(2) =>\(y^2-5x^2=4\) (3)

(1) sẽ trở thành: \(x^3 + (y^2 - 5x^2)y = y^3 + 16x\)

=>\(x^3 - 5x^2y - 16x = 0\)

=>\(x(x^2 - 5xy - 16) = 0\)

TH1: x=0

=>\(y^2=4+5x^2=4\)

=>y=2 hoặc y=-2

TH2: x<>0 và \(x^2-5xy-16=0\)

=>\(x^2-5xy=16\)

\(y^2-5x^2=4\)

=>\(-20x^2+4y^2=16\)

=>\(-20x^2+4y^2=x^2-5xy\)

=>\(-21x^2+5xy+4y^2=0\)

=>\(21x^2-5xy-4y^2=0\)

=>\(21x^2-12xy+7xy-4y^2=0\)

=>3x(7x-4y)+y(7x-4y)=0

=>(7x-4y)(3x+y)=0

TH1: 3x+y=0

=>y=-3x

Thay y=-3x vào (3), ta được:

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

=>\(9x^2-5x^2=4\)

=>\(4x^2=4\)

=>\(x^2=1\implies x=\pm1\)

x=1

=>y=-3

x=-1

=>y=3

TH2: 7x-4y=0

=>7x=4y

=>x=4/7y

=>y=7/4x

Thay y=7/4x vào (3), ta được:

\(\left(\frac{7}{4}x\right)^2-5x^2=4\iff\frac{49}{16}x^2-5x^2=4\iff-\frac{31}{16}x^2=4\quad\)

=>x∈∅

Yêu cầu đề là gì vậy bạn?

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

$=3x^2+3x-3x^2-2x$

$=x$

$2x(x^2-5x+6)+(x-1)(x+3)$

$=2x^3-10x^2+12x+x^2+2x-3$

$=2x^3-9x^2+14x-3$

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=-xy-y^2+4y\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=-y\left(x+y-4\right)\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)

Thế trên xuống dưới:

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

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow x^2+1=0\left(l\right)\\\left(x+y-4\right)\left(x+y-2\right)=-1\left(1\right)\end{matrix}\right.\)

Xét (1), đặt \(x+y-4=t\)

\(\Rightarrow t\left(t+2\right)=-1\Leftrightarrow\left(t+1\right)^2=0\Rightarrow t=-1\)

\(\Rightarrow x+y-4=-1\Rightarrow y=3-x\)

Thế vào pt đầu:

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

\(\Leftrightarrow...\)

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. 

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

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

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

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