(\(\left(6x+1\right)\times\left(6x-1\right)\) )\(^2\)

bi nham roi

a/ (6x+1)2+(6x-1)²-2(1+6x)(6x-1)

=36x²+12x+1+36x²-12x+1-2(6x-1+36x²-6x)

=36x²+12x+1+36x²-12x+1+2-72x²

=1+1+2=4

b/ 3(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)

Ta có: 3=4-1=2²-1 

<=> (2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)

  =(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)

  =(2⁸-1)(2⁸+1)(2¹⁶+1)

  =(2¹⁶-1)(2¹⁶+1)

  =2³²-1

mình biết làm bài này

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

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

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

$=2^2$

$=4$

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

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

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

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

$=2^{32}-1$

a) \(=\left[\left(6x+1\right)+\left(6x-1\right)\right]^2\)

\(=\left(12x\right)^2\)

\(=144x^2\)

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

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

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

$=2^2$

$=4$

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

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

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

$=3(2^2-1)\cdot\dfrac{2^{32}-1}{2^4-1}$

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

$=\dfrac{3(2^{32}-1)}{5}$

a, Áp dụng hằng đẳng thức số 2 ta có

[6x+1-(6x+1)]²=(6x+1-6x-1)²=0

b, (2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)

=(2⁸-1)(2⁸+1)(2¹⁶+1)

=(2¹⁶-1)(2¹⁶+1)=2³²-1

Tao không biết

Lưu lê thanh hạ rảnh lên à bạn ???

\(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1+6x-1\right)^2\)

\(=36x^2\)

a)\(\left(6x+1\right)^2+\left(6x+1\right)^2-2\left(1-6x\right)\left(6x-1\right)\)

\(\Leftrightarrow36x^2+1+36x^2+1-2\left(36x^2-1\right)\)

\(\Leftrightarrow1+1+2=4\)

b) \(\Leftrightarrow4x^2-9-\left(x^2+10x+25\right)-\left(x^2+x-2\right)\)

\(\Leftrightarrow2x^2-11x-32\)

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3x^2+3x+3x+3}\)

\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3x\left(x+1\right)+3\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{\left(3x+3\right)\left(x+1\right)}\)

\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)^2}=\frac{x^2+1}{3\left(x+1\right)}\)

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2.\left(x+1\right)+\left(x+1\right)}{3.\left(x^2+2x+1\right)}=\frac{\left(x+1\right)\left(x^2+1\right)}{3.\left(x+1\right)^2}=\frac{x^2+1}{3.\left(x+1\right)}\)