\(\left(x^2+7x+12\right)\left(x^2-15x+56\right)=180\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(x+4\right)\left(x-7\right)\left(x-8\right)-180=0\)

\(\Leftrightarrow\)\(\left(x^2-4x-21\right)\left(x^2-4x-32\right)-180=0\)

Đặt     \(x^2-4x-21=t\)  ta có:

                         \(t\left(t-11\right)-180=0\)

           \(\Leftrightarrow\)\(t^2-11t-180=0\)

           \(\Leftrightarrow\)\(t^2-20t+9t-180=0\)

           \(\Leftrightarrow\)\(\left(t-20\right)\left(t+9\right)=0\)

           \(\Leftrightarrow\)\(\orbr{\begin{cases}t-20=0\\t+9=0\end{cases}}\)

  P/S:đến đây bn thay trở lại rồi tìm   x   nhé! chúc bn hok tốt

1/ (x-63)(x+10)(4x² -188x-2520)

2/ 9(x-1)(2x-1)(64x² + 208x+32)/8

3/ (x+2)(x-6)(x² -4x -41)

\(=\left(x^2-11x+26+4x-14\right)\left(x^2-11x+26-4x+14\right)+16\left(x^2-7x+12\right)-60\)

\(=\left(x^2-11x+26\right)^2-\left(4x-14\right)^2+\left(16x^2-2\cdot4\cdot14x+14^2\right)-64\)

\(=\left(x^2-11x+18\right)\left(x^2-11x+34\right)-\left(4x+14\right)^2+\left(4x+14\right)^2\)

\(=\left(x^2-2x-9x+18\right)\left(x^2-11x+34\right)\)

\(=\left(x-2\right)\left(x-9\right)\left(x^2-11x+34\right)\)

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\) \(\frac{1}{x^2+15x+56}=\frac{1}{14}\)

<=>\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)+...+ \(\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}\)= \(\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

<=> \(\frac{x+8-x-1}{\left(x+1\right)\left(x+8\right)}=\frac{1}{14}\)

<=>\(\frac{7.14}{14\left(x+1\right)\left(x+8\right)}=\frac{\left(x+1\right)\left(x+8\right)}{14\left(x+1\right)\left(x+8\right)}\)

<=> \(x^2+9x+8=98\)<=> \(x^2+9x-90=0\)

<=> (x-6)(x+15) =0 

<=> \(\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)

Vậy phương trình có 2 nghiệm  x  \(\in\left(6,15\right)\)

==============

- Do ko biết viết dấu ngoặc nhọn nên thay = dấu ngoặc tròn

- Đề ko rõ ràng , lần sau nhớ ghi yêu cầu ?  

2x^2 + x - 6

= 2x^2 + 4x - 3x - 6

= 2x(x + 2) - 3(x + 2)

= (2x - 3)(x + 2)

7x^2 + 50x + 7 

= 7x^2 + x + 49x + 7

= 7x(x + 7) + x + 7

= (7x + 1)(x + 7)

12x^2 + 7x - 12

15x^2 +  7x - 2

= 15x^2 - 3x + 10x - 2

= 3x(5x - 1) + 2(5x - 1) 

= (3x + 2)(5x - 1)

a^2 - 5a - 14

= a^2 + 2a - 7a - 14

= a(a + 2) - 7(a + 2)

= (a - 7)(a + 2)

2x^2 + 5x + 2

= 2x^2 + x + 4x + 2

= 2x(x + 2) + x + 2

= (2x + 1)(x + 2)

\(2x^2+x-6=2x^2+4x-3x-6\)

\(=2x\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(2x-3\right)\)

\(7x^2+50x+7\)

\(=7x^2+x+49x+7\)

\(=x\left(7x+1\right)+7\left(7x+1\right)\)

\(=\left(7x+1\right)\left(x+7\right)\)

\(12x^2+7x-12\)

\(=12x^2+16x-9x-12\)

\(=4x\left(3x+4\right)-3\left(3x+4\right)\)

\(=\left(3x+4\right)\left(4x-3\right)\)

\(4p^2-36p+56\)

\(=4p^2+8p+28p+56\)

\(=4p\left(p+2\right)+28\left(p+2\right)\)

\(=\left(p+2\right)\left(4p+28\right)\)

ĐKXĐ : Tự tìm nha : )

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

=> \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{x+8}{\left(x+1\right)\left(x+8\right)}-\frac{x+1}{\left(x+8\right)\left(x+1\right)}=\frac{1}{14}\)

=> \(14\left(x+8-x-1\right)=\left(x+1\right)\left(x+8\right)\)

=> \(x^2+x+8x+8=98\)

=> \(x^2+9x-90=0\)

=> \(\left(x+15\right)\left(x-6\right)=0\)

=> \(\left[{}\begin{matrix}x+15=0\\x-6=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-15\\x=6\end{matrix}\right.\) ( TM )

Vậy phương trình trên có nghiệm là \(S=\left\{6,-15\right\}\)

Lời giải:

PT \(\Leftrightarrow \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+....+\frac{1}{(x+7)(x+8)}=\frac{1}{14}\)

(ĐK: $x\neq -1;-2;...;-8$)

\(\Leftrightarrow \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+....+\frac{(x+8)-(x+7)}{(x+7)(x+8)}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\Leftrightarrow \frac{7}{x^2+9x+8}=\frac{1}{14}\)

\(\Rightarrow x^2+9x+8=98\Leftrightarrow x^2+9x-90=0\Rightarrow x=6\) hoặc $x=-15$ (đều thỏa mãn)

Vậy........