a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

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

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

a) (2x - 17 )⁵ = (2x -17 )⁵

=> Với mọi x \(\in\)N thì : ( 2x - 17 )⁵ = (2x -17 )⁵

Vậy :...

b) 

=> Với x = 0 hoặc x = 1 thì  5^x = 3^x

Vậy :...

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

A min khi 2x+1=-1

=>x=-1

120 dm² x 5 + 4m² = 600 dm² +  4 m² = 6 m² + 4m² = 10 m²

\(2^{x+3}.2=2^2.3+52\)

\(=>2^{x+3}.2=64\)

\(=>2^{x+3}=64:2\)

\(=>2^{x+3}=32\)

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

=>x+3=5

=>x=5-3

=>x=2

Vậy ...........

 2^{x + 3} . 2 = 2² . 3 + 52

 2^{x + 3} . 2 = 4 . 3 + 52

 2^{x + 3} . 2 = 12 + 52

 2^{x + 3} . 2 = 64

     2^{x + 3}  = 64 : 2

     2^{x + 3}  = 32

     2^{x + 3} = 2⁵

     x + 3 = 5

           x = 5 - 3

           x = 2

Vậy x = 2

Ta có 

20x5 chia hết cho 3

=> 2 + 0 + x  + 5 chia hết cho 3

=> 7 + x chia hết cho 3

mà x là chữ số

=> x thuộc { 2 ; 5 ; 8 }

Vì x:20x5 chia hết cho 3

=>x:20x5=3k(k thuộc Z)

=>x:4=3k

=>x=3k.4

=>x=12k.

Vậy x=12k khi  x:20x5 chia hết cho 3.

ĐKXĐ: x^2-4<>0

=>\(x\notin\left\{2;-2\right\}\)

chọn D