a) A = -1;                        b) B = ( x   +   y ) 3  =1.

\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3x^2-6xy-3y^2\)
\(A=4x^2+4xy+4y^2-3x^2-6xy-3y^2\)
\(A=x^2-2xy+y^2\)
\(A=\left(x-y\right)^2\)
\(A=2^2=4\)

A=2(x³-y³)-3(x+y)²

A=2(x-y)(x²+xy+y²)-3(x²+2xy+y²)

A=2.2(x²+xy+y²)-3(x²+2xy+y²)

A=4(x²+xy+y²)-3x²+6xy+3y²

A=4x²+4xy+y²-3x²-6xy+3y²

A=x²-2xy+y²

A=(x-y)²

A= 2²

A=4

?????

bạn sửa dùm mình câu A=4(x2+xy+y2)-3x2+6xy+3y2  là -6xy-3y² nha. Mình cảm ơn 

`2(x^3+y^3)-3(x^2+y^2)+100`

`=2(x+y)(x^2-xy+y^2)-3x^2-3y^2+100`

`=2x^2-2xy+2y^2-3x^2-3y^2+100`

`=-x^2-2xy-y^2+100`

`=-(x+y)^2+100`

`=-1+100=99`

a) \(A=x^3+y^3+3xy\)

\(=x^3+y^3+3xy\left(x+y\right)\) (do \(x+y=1\))

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

\(=\left(x+y\right)^3\) \(=1\)

b) \(B=x^3-y^3-3xy\)

\(=x^3-y^3-3xy\left(x-y\right)\) (do \(x-y=1\))

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

\(=\left(x-y\right)^3\) \(=1\)

\(A=2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]-3\left[\left(x-y\right)^2+4xy\right]\\ A=2\left(8+6xy\right)-3\left(4+4xy\right)\\ A=12xy+16-12-12xy=4\)

Vậy chọn A

(x - 5)² = (3 + 2x)²

(x - 5)² - (3 + 2x)² = 0

[(x - 5) - (3 + 2x)][(x - 5) + (3 + 2x)] = 0

(x - 5 - 3 - 2x)(x - 5 + 3 + 2x) = 0

(-x - 8)(3x - 2) = 0

-x - 8 = 0 hoặc 3x - 2 = 0

*) -x - 8 = 0

-x = 8

x = -8

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = -8; x = 2/3

--------------------

27x³ - 54x² + 36x = 9

27x³ - 54x² + 36x - 9 = 0

27x³ - 27x² - 27x² + 27x + 9x - 9 = 0

(27x³ - 27x²) - (27x² - 27x) + (9x - 9) = 0

27x²(x - 1) - 27x(x - 1) + 9(x - 1) = 0

(x - 1)(27x² - 27x + 9) = 0

x - 1 = 0 hoặc 27x² - 27x + 9 = 0

*) x - 1 = 0

x = 1

*) 27x² - 27x + 9 = 0

Ta có:

27x² - 27x + 9

= 27(x² - x + 1/3)

= 27(x² - 2.x.1/2 + 1/4 + 1/12)

= 27[(x - 1/2)² + 1/12] > 0 với mọi x ∈ R

⇒ 27x² - 27x + 9 = 0 (vô lí)

Vậy x = 1

A = x² + y²

= x² - 2xy + y² + 2xy

= (x - y)² + 2xy

= 4² + 2.1

= 16 + 2

= 18

B = x³ - y³

= (x - y)(x² + xy + y²)

= (x - y)(x² - 2xy + y² + xy + 2xy)

= (x - y)[(x - y)² + 3xy]

= 4.(4² + 3.1)

= 4.(16 + 3)

= 4.19

= 76

C = x⁴ + y⁴

= (x²)² + (y²)²

= (x²)² + 2x²y² + (y²)² - 2x²y²

= (x² + y²)² - 2x²y²

= (x² - 2x²y² + y² + 2x²y²)² - 2x²y²

= [(x - y)² + 2x²y²]² - 2x²y²

= (4² + 2.1²)² - 2.1²

= (16 + 2)² - 2

= 18² - 2

= 324 - 2

= 322

a: =>(2x+3)^2-(x-5)^2=0

=>(2x+3+x-5)(2x+3-x+5)=0

=>(x+8)(3x-2)=0

=>x=2/3 hoặc x=-8

b: =>27x^3-54x^2-36x-9=0

=>3x^3-6x^2-4x-1=0

=>\(x\simeq2,57\)

c: A=x^2+y^2=(x-y)^2+2xy=4^2+2=18

B=x^3-y^3=(x-y)^3+3xy(x-y)

=4^3+3*1*4

=64+12=76

C=(x^2+y^2)^2-2x^2y^2

=18^2-2*1^2=322

\(a,A=x^2+y^2\\=x^2-2xy+y^2+2xy\\=(x-y)^2+2xy\\=2^2+2\cdot1\\=4+2\\=6\)

\(b,x+y=1\\\Leftrightarrow (x+y)^3=1^3\\\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\\\Leftrightarrow x^3+3xy(x+y)+y^3=1\\\Leftrightarrow x^3+3xy\cdot1+y^3=1\\\Rightarrow A=1\)

a) Ta có:

\(x-y=2\)

\(\Rightarrow\left(x-y\right)^2=2^2\)

\(\Rightarrow x^2-2xy+y^2=4\)

Mà: \(xy=1\)

\(\Rightarrow\left(x^2+y^2\right)-2\cdot1=4\)

\(\Rightarrow x^2+y^2=4+2\)

\(\Rightarrow x^2+y^2=6\)

b) Ta có: 

\(x+y=1\)

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

\(\Rightarrow x^3+3x^2y+3xy+y^3=1\)

\(\Rightarrow x^3+3xy\left(x+y\right)+y^3=1\) 

Mà: x + y = 1

\(\Rightarrow x^3+3xy\cdot1+y^3=1\)

\(\Rightarrow x^3+3xy+y^3=1\)