\(\Leftrightarrow\left(36x^2+84x+48\right)\left(36x^2+84x+49\right)=72\)

\(\Leftrightarrow t\left(t+1\right)=72\) ( với \(t=36x^2+84x+48\) )

\(\Leftrightarrow t^2+t-72=0\Leftrightarrow\left(t-8\right)\left(t+9\right)=0\)

\(\Leftrightarrow t-8=0\) ( do \(t+9=36x^2+84x+49+8=\left(6x+7\right)^2+8>0\forall x\))

\(\Leftrightarrow36x^2+84x+48=8\)

\(\Leftrightarrow\left(6x+7\right)^2=9\Leftrightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{matrix}\right.\) ( TM )

x=\(\dfrac{-2}{3}\)

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 6x+7=a Ta có \(\left(a^2-1\right)a^2=72\Leftrightarrow a^4-a^2-72=0\Leftrightarrow\left(a^2+8\right)\left(a^2-9\right)=0\)Mà a^2+8>0 nên \(a^2-9=0\Rightarrow a=+-3\Rightarrow6x+7=+-3\Rightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{matrix}\right.\)

Ta có : \(\left(6x+6\right)\left(6x+8\right)\left(6x+7\right)^2=72\)

=> \(\left(36x^2+84x+48\right)\left(36x^2+84x+49\right)=72\)

- Đặt \(36x^2+84x+48=a\) ta được phương trình :

\(a\left(a+1\right)=72\)

=> \(a^2+a-72=0\)

=> \(\left(a-8\right)\left(a+9\right)=0\)

=> \(\left[{}\begin{matrix}a=8\\a=-9\end{matrix}\right.\)

- Thay lại \(36x^2+84x+48=a\) vào phương trình trên ta được :

\(\left[{}\begin{matrix}36x^2+84x+48=8\\36x^2+84x+48=-9\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(6x+7\right)^2=9\\\left(6x+7\right)^2=-8\left(vl\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}6x+7=\sqrt{9}\\6x+7=-\sqrt{9}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}6x=-4\\6x=-10\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{2}{3};-\frac{5}{3}\right\}\)

https://olm.vn/hoi-dap/detail/68292664726.html

Đặt

6x+7 = 7 , ta có

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

\(\Rightarrow t^4-t^2-72=0\)

Lại đặt \(t^2=a\) (a \(\ge0\) )

\(\Rightarrow a^2-a-72=0\Rightarrow\left(a+8\right)\left(a-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-8\left(ktm\right)\\a=9\left(tm\right)\end{matrix}\right.\)

a = 9 => \(\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\)

Với t = 3

=> 6x + 7 =3

=> 6x = -4

=> x= \(-\frac{2}{3}\)

Với t = -3

=> 6x + 7 = -3

=> 6x = -10

=> x = \(-\frac{5}{3}\)

Vậy.....

b)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

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

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

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\Rightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+7\right)\left(x+4\right)}=\frac{1}{18}\Rightarrow x^2+11x+28-54=0\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

a) Ta có:

(6x+8)(6x+6)(6x+7)² = 72

Đặt \(6x+7=a\)

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

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

\(\Leftrightarrow\left(a^4+8a^2\right)+\left(-9a^2-72\right)=0\)

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

Đễ thấy \(a^2+8>0\)

\(\Rightarrow a^2-9=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x+7=3\\6x+7=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)

b)

Đặt \(6x+7=a\)

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

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

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

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

\(\Leftrightarrow a^4+8a^2-9a^2-72=0\)

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

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

Vì \(a^2+8>0\)

\(\Rightarrow a^2-9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}6x=-4\\6x=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{5}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-\frac{2}{3};-\frac{5}{3}\right\}\)

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}

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

\(x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0\)

\(\Leftrightarrow x^6-x^5-5x^5+5x^4+10x^4-10x^3-10x^3+10x^2+5x^2-5x-x+1=0\)

\(\Leftrightarrow x^5\left(x-1\right)-5x^4\left(x-1\right)+10x^3\left(x-1\right)-10x^2\left(x-1\right)+5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^5-5x^4+10x^3-10x^2+5x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^5-x^4-4x^4+4x^3+6x^3-6x^2-4x^2+4x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x-1\right)-4x^3\left(x-1\right)+6x^2\left(x-1\right)-4x\left(x-1\right)+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-4x^3+6x^2-4x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-x^3-3x^3+3x^2+3x^2-3x-x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3\left[x^3-3x^2+3x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3\left[x^3-x^2-2x^2+2x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^4\left[x^2-2x+1\right]=0\Leftrightarrow\left(x-1\right)^6=0\Leftrightarrow x=1\)