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: \(x\ne-4;-5;-6;-7\)

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

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

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

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

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

Mày nhìn cái chóa j

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

ĐKXĐ: x khác -4;-5;-6;-7

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

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

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

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

\(\Rightarrow\frac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow3.18=x^2+11x+28\)

\(\Rightarrow x^2+11x-26=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}\left(tm\right)}\)

Vậy...

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

a. \(x^2+9x+20=\left(x^2+4x\right)+\left(5x+20\right)\)

\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)

Tương tự: \(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

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

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

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

\(=18\left(x+7\right)-18\left(x+4\right)=\left(x+7\right)\left(x+4\right)\)

\(=x^2+11x+28=54\)

\(=x^2+11x-26=0\)

\(=\left(x^2-2x\right)+\left(13x-26\right)=0\)

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

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

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

b. \(\left(3x-2\right)\left(4x+5\right)=0\)

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

hic, mk chx học