Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)

\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)

Dấu "=" \(x=y=\dfrac{1}{2}\)

Ta có \(x-y=1\)

\(=>x+y=\left(x+y\right).\left(x-y\right)\)
\(A=\left(x+y\right).\left(x-y\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)

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

\(A=\left(x^4-y^4\right).\left(x^4+y^4\right)\)

\(A=x^8-y^8\)

= \(-\left[\left(x-y\right)\left(x^2-y^2\right)\left(x^4-y^4\right)\left(x^8-y^8\right)\left(x^{16}-y^{16}\right)\right]\)

= \(-\left[\left(x-y\right)\left(x-y\right)^2\left(x-y\right)^4\left(x-y\right)^8\left(x-y\right)^{16}\right]\)

= \(-\left(1\cdot1^2\cdot1^4\cdot1^8\cdot1^{16}\right)\)

= -1

1: =(x+y-3x)(x+y+3x)

=(-2x+y)(4x+y)

2: =(3x-1-4)(3x-1+4)

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

=3(x+1)(3x-5)

3: =(2x)^2-(x^2+1)^2

=-[(x^2+1)^2-(2x)^2]

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

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

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

=(5x-5y-4x-4y)(5x-5y+4x+4y)

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

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

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

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)

a. Do \(x=y-1\Rightarrow x-y=1\)

Ta có:

\(A=x^3-y^3-3xy=\left(x-y\right)^3+3xy\left(x-y\right)-3xy=1^3+3xy.1-3xy=1\left(đpcm\right)\)

b. \(B=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)

(Do \(x-y=1\))

(Bạn áp dụng hằng đẳng thức \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)vào bài toán)

Kết quả, \(B=x^{16}-y^{16}\left(đpcm\right)\)

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

Hay x³- 3xy(x-y) -  y³=1  => x³- y³ -3xy =1

b) 1.(x+y)(x²+y²)(x⁴+y⁴)(x⁸+y⁸) = (x-y)(x+y)......................=(x²-y²)(x²+y²)..........=(x^4-y⁴)(x⁴+y⁴)......=(x⁸-y⁸)(x⁸+y⁸) =x¹⁶-y¹⁶

Bài 1 :

a ) Ta có :

\(\left(x+y\right)^2=x^2+y^2+2xy=20+16=36\)

b ) Ta có :

\(x^2+y^2=\left(x+y\right)^2-2xy=64-30=34\)

Câu 8: D

Câu 9: B