đặt 6x+7=a

suy ra (a-1)(a+1)a²=72

(a²-1)a²=72

a⁴-a²+1/4=289/4

(a²-1/2)=289/4

hoặc a²-1/2=17/2

         a²-1/2=-17/2

suy ra hoặc a²=9

                    a²=-8(loại vì a²>=0>-8 với mọi a )

suy ra a=3

            a=-3

hay 6x+7=3 suy ra x=-2/3

       6x+7=-3 suy ra x=-5/3

vậy S={-2/3,-5/3}

đặt t = x + 1. Phương trình có dạng:

(6t + 1)².t .(3t +1) = 6

<=> (36t² + 12t + 1).(3t² + t)  = 6

<=> [12.(3t² + t) + 1](3t² + 1) = 6

<=> 12.(3t² +1)² + (3t² +1) - 6 = 0

<=> 12.(3t² +1)² + 9(3t² +1) - 8.(3t² +t)  - 6 = 0

<=> 3(3t² + t). [4(3t² +t) +3] - 2. [4(3t² +t) +3] = 0

<=> [4(3t² +t) +3]. [3(3t² +t) - 2] = 0

<=> 4(3t² +t) +3 = 0 hoặc 3(3t² +t) - 2 = 0

+)  4(3t² +t) +3 = 0 <=> 12t² + 4t + 3 = 0  Vô nghiệm vì 12t² + 4t + 3 = 8t² + (2t +1)² + 2 > 0 với mọi t

+) 3(3t² +t) - 2 = 0 <=> 9t² + 3t - 2 = 0 <=> 9t² + 6t - 3t - 2 = 0 <=> (3t + 2)(3t -1) = 0

=> t = -2/3 hoặc t = 1/3

=> x + 1 = -2/3 hoặc x + 1 = 1/3

=> x = -5/3 hoặc x = -2/3

a/ Đặt \(6x+7=a\Rightarrow\left\{{}\begin{matrix}6x+8=a+1\\6x+6=a-1\end{matrix}\right.\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)a^2-72=0\)

\(\Leftrightarrow\left(a^2-1\right)a^2-72=0\)

\(\Leftrightarrow a^4-a^2-72=0\)

\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)

\(\Leftrightarrow a^2=9\) (do \(a^2+8>0\))

\(\Rightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne-4;-5;-6;-7\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

Đặt \(x^2+6x+7=t\)

Bài toán trở thành tìm m để phương trình: \(\left(t-2\right)\left(t+1\right)=m-1\) (1) có nghiệm \(t< 0\)

\(\left(1\right)\Leftrightarrow t^2-t-1=m\)

Xét hàm \(f\left(t\right)=t^2-t-1\)

\(f\left(0\right)=-1\) và hàm số nghịch biến khi \(t< 0\)

\(\Rightarrow f\left(t\right)>-1\) \(\forall t< 0\)

\(\Rightarrow\) phương trình \(f\left(t\right)=m\) có nghiệm \(t< 0\) khi và chỉ khi \(m>-1\)

Vậy với \(m>-1\) thì pt đã cho có nghiệm thỏa \(x^2+6x+7< 0\)

a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\\ =\left[\left(6x\right)^2+2\cdot6x+1^2\right]+\left[\left(6x\right)^2-2\cdot6x\cdot1+1^2\right]-2\left[\left(6x\right)^2-1^2\right]\\ =36x^2+12x+1+36x^2-12x+1-72x^2-2\\ =0\)b)\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^{16}-1\right)\left(2^{16}+1\right)\\ =2^{32}-1\)

a)  (6x + 1)² + (6x - 1)² - 2(1 + 6x)(6x - 1) 

= (6x + 1 - 6x + 1)² = 4

b) 3(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ +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

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

\(=\left(6x+1-6x+1\right)^2=2^2=4\)

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

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

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

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

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