6x:(16-2mũ2)=7mũ3:7mũ2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi A =2^0+2^1+2^2+...+2^2015=1+2^1+2^2+...+2^2015
Ax2=(1+2^1+2^2+...+2^2015)x2
Ax2=2^1+2^2+2^3+...+2^2016
Ax2-A=2^1+2^2+2^3+...+2^2016-(2^0+2^1+2^2+...+2^2015)
A=2^1+2^2+2^3+...+2^2016-1-2^1-2^2-...-2^2015
A=2^2016-1
Vậy A=2^2016-1
Ai ấn Đúng 0 sẽ may mắn cả năm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
6x=-16+2x
=>6x-2x=-16
=>4x=-16
=>x=-4
Vậy x=4.
2(x-7)=-16
=>2x-14=-16
=>2x=-16+14
=>2x=-2
=>x=-1
Vậy x=-1.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a) }\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\\ =\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\\ =\left[6x+1-6x+1\right]^2\\ =2^2\\ =4\)
\(\text{b) }3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^{16}-1\right)\left(2^{16}+1\right)\\ =2^{32}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1\right)-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left[\left(6x-1\right)-2\left(1+6x\right)\right]\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1-2-12x\right)\)
\(=36x^2+12x+1+\left(6x-1\right)\left(-6x-3\right)\)
\(=36x^2+12x+1+\left(-36x^2-12x+3\right)\)
\(=36x^2+12x+1-36x^2-12x+3\)
\(=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a.
$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2=2^2=4$
b.
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$
c.
$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$
$\Rightarrow C=\frac{5^{32}-1}{2}$