\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

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

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

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Đặt : \(P=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

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

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

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

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

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

1,

Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

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

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

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

\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(A=2^{32}-1\)

Vậy \(A=2^{32}-1\)

2, \(x^2-6x=-9\)

\(x^2-6x+9=0\)

\(\left(x-3\right)^2=0\)

\(x-3=0\)

\(x=3\)

Vậy \(x=3\)

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

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

$=[(6x+1)-(6x-1)]^2$

$=2^2$

$=4$

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$

$=3\cdot\dfrac{2^{32}-1}{2^2-1}$

$=3\cdot\dfrac{2^{32}-1}{3}$

$=2^{32}-1$

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

$=2x^3-3x-5x^3-x^2+x^2$

$=2x^3-5x^3-3x$

$=-3x^3-3x$

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

$3x(x-2)-5x(1-x)-8(x^2-3)$

$=3x^2-6x-5x+5x^2-8x^2+24$

$=3x^2+5x^2-8x^2-11x+24$

$=-11x+24$

$=24-11x$

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

C = (2x-3)²-(x+4)(2x-1) -(x+3)²

(Chuyển đổi các hằng đẳng thức)

    = (4x²-12x+9)-(2x²-x+8x-4)-(x²+6x+9)

    =  4x²-12x+9-2x²+x-8x+4-x²-6x-9

(Ta thu gọn các hạng tử đồng dạng với nhau)

    = x²-25x-14 

các bn giúp mình nhé

link câu a nek https://olm.vn/hoi-dap/detail/8804510287.html

$x^4-12x^3+12x^2-12x+111$ tại $x=11$

$=11^4-12\cdot11^3+12\cdot11^2-12\cdot11+111$

$=14641-15972+1452-132+111$

$=100$

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

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

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

$=36x^2+12x+1+3(36x^2-12x+1)$

$=36x^2+12x+1+108x^2-36x+3$

$=144x^2-24x+4$

$=4(36x^2-6x+1)$

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$

$=3\cdot\dfrac{2^{32}-1}{2^2-1}$

$=3\cdot\dfrac{2^{32}-1}{3}$

$=2^{32}-1$

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!

$(x-2y)^2$

$=x^2-2\cdot x\cdot 2y+(2y)^2$

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

$(2x^2+3)^2$

$=(2x^2)^2+2\cdot2x^2\cdot3+3^2$

$=4x^4+12x^2+9$

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

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

$=x^3+2x^2+4x-2x^2-4x-8$

$=x^3-8$

$(2x-1)^3$

$=(2x)^3-3(2x)^2\cdot1+3\cdot2x\cdot1^2-1^3$

$=8x^3-12x^2+6x-1$