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)

a) Ta có: \(x^2-x-12=0\)

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

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

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

Lời giải:

\(P(x)=x(x+2)(x+3)(x+5)-7\)

\(=[x(x+5)][(x+2)(x+3)]-7\)

\(=(x^2+5x)(x^2+5x+6)-7\)

\(=a(a+6)-7\) (đặt \(x^2+5x=a\) )

\(=a^2+6a-7=a^2-a+7a-7\)

\(=a(a-1)+7(a-1)=(a-1)(a+7)\)

\(=(x^2+5x-1)(x^2+5x+7)\)

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\(Q(x)=(4x-2)(10x+4)(5x+7)(2x+1)+17\)

\(=4(2x-1)(5x+2)(5x+7)(2x+1)+17\)

\(=4[(2x-1)(5x+7)][(5x+2)(2x+1)]+17\)

\(=4(10x^2+9x-7)(10x^2+9x+2)+17\)

\(=4a(a+9)+17\) (đặt \(10x^2+9x-7=a\)

\(=4a^2+36a+17=(2a+9)^2-8^2\)

\(=(2a+9-8)(2a+9+8)=(2a+1)(2a+17)\)

\(=(20x^2+18x-13)(20x^2+18x+3)\)

\(R(x)=(3x+2)(3x-5)(x-1)(9x+10)+24x^2\)

\(=[(3x+2)(3x-5)][(x-1)(9x+10)]+24x^2\)

\(=(9x^2-9x-10)(9x^2+x-10)+24x^2\)

\(=(a-9x)(a+x)+24x^2\) (đặt \(9x^2-10=a\) )

\(=a^2-8ax+15x^2=(a^2-5ax)-(3ax-15x^2)\)

\(=a(a-5x)-3x(a-5x)=(a-3x)(a-5x)\)

\(=(9x^2-3x-10)(9x^2-5x-10)\)

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\(H(x)=(x-18)(x-7)(x+35)(x+90)-67x^2\)

\(=[(x-18)(x+35)][(x-7)(x+90)]-67x^2\)

\(=(x^2+17x-630)(x^2+83x-630)-67x^2\)

\(=a(a+66x)-67x^2\) (đặt \(x^2+17x-630=a\) )

\(=a^2-ax+67ax-67x^2\)

\(=a(a-x)+67x(a-x)=(a-x)(a+67x)\)

\(=(x^2+16x-630)(x^2+84x-630)\)

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

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

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

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

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  \(\in\left(6,15\right)\)

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

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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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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