x2 là x² hả bạn???

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: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)

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

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)

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 $(a-b)(a^2+ab+b^2)=a^3-b^3$:

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

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

$=x^3y^3$

1: Đặt S=x+y; P=xy

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

=>\(3(S^2-3P)=7S\)

=>\(9P=3S^2-7S\)

=>\(P = \frac{3S^2 - 7S}{9}\)

x,y là các số nguyên

=>\(\left(x+y\right)^2\ge4xy\)

=>\(S^2\ge4P\)

=>\(S^2-4P\ge0\)

=>\(S^2-4\cdot\frac{3S^2 - 7S}{9}\ge0\)

=>\(9S^2-12S^2+28S\ge0\)

=>\(-3S^2+28S\ge0\)

=>0<=S<=28/3

S∈Z; 9P=S(3S-7)

=>S⋮9

=>S=0 hoặc S=9

TH1: S=0

\(P=\frac{3\cdot0^2-7\cdot0}{9}=0\)

Do đó, ta có: x+y=0 và xy=0

=>x=y=0

TH2: S=9

=>\(P=\frac{3\cdot9^2-7\cdot9}{9}=\frac{3\cdot81-63}{9}=\frac{9\cdot\left(3\cdot9-7\right)}{9}=20\)

Do đó: x+y=9 và xy=20

=>(x;y)∈{(4;5);(5;4)}

a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

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

\(=9y^3\)

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

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

\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)

\(=x^3-x^2-x+1-x^3-8\)

\(=-x^2-x-7\)

\(b,\left(x+2\right)^2-25\)

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

\(=\left(x-3\right)\left(x+7\right)\)

\(c,36\left(x-y\right)^2\)

\(=36\left(x^2-2xy+y^2\right)\)

\(=36x^2-72xy+36y^2\)

\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)

\(=\left(x+\dfrac{1}{4}\right)^2\)

\(e,2x^4y^3-3x^2y^4+5x^3y^4\)

\(=x^2y^3\left(2x^2-3y+5xy\right)\)

Các câu còn lại làm tương tự, chú ý sd HĐT

a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)

b: \(=-5y-9+xy\)

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

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

$=-4x-12+x^2-3x+1$

$A=x^2-7x-11$

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

$=4(x^2+2x+1)-4x^2-8x$

$B=4$